Percent ConcentrationBy: Cristel ImbagThe use of percentages is a common way of expressing the concentration of a solution. It is a straightforward approach that you have used earlier when dealing with the composition of compounds. There are, however, some differences. One is that the concentrations of solution are variable while the composition of compounds is constant. Another is that the percentages can be calculated using volumes as well as weights, or even both together. One way of expressing concentrations, with which you might be familiar, is by volume percent. Another is by weight percent. Still another is a hybrid called weight/volume percent. In the pages of this section we will look at how to calculate each and the cases where each is generally used.
Mass Percentage One way
of representing the concentration of an element in a compound or a
component in a mixture. Mass percentage is calculated as the mass of a
component divided by the total mass of the mixture, multiplied by 100%. Volume Percentage Volume percent is a common expression of a solution's concentration. It is defined as: It is related to volume fraction by dividing by 100%. Volume percent is usually used when the solution is made by mixing twofluids, such as liquids or gases. However, percentages are only additive forideal gases. The solute is what you get from the solution. For example, if you distilled salt water, the solution would be the salt water, and the solvent would be thedistilled water. A solution consists of a solvent, which is commonly a liquid, and a solute, which may be any substance that dissolves in the solvent. For example, if salt water (a solution) is distilled, water (the solvent) passes over into the receiver and salt (the solute) remains in the still. In the case of a mixture of ethanol and water, which are miscible in all proportions, the designation of solvent and solute is arbitrary. The volume of such a mixture is slightly less than the sum of the volumes of the components. Thus, by the above definition, the term "40% alcohol by volume" refers to a mixture of 40 volume units of ethanol with enough water to make a final volume of 100 units, rather than a mixture of 40 units of ethanol with 60 units of water. Example: You have 40.0ml of C6H5CH3 that is put into 75.0ml of C6H6, what is the percent volume of C6H5CH3? Solution: (40.0ml) / (115.0ml) x 100 = 34.8% of C6H5CH3
Weight/Volume Percentage * weight/volume percentage concentration is usually abbreviated as w/v (%) * w/v (%) = mass solute ÷ volume solution x 100 * common units are g/100mL (%) * weight/volume is a useful concentration measure when dispensing reagents. Reference:About.com Chemistry Volume-volume percentage, Chembuddy websiteAusetute.com.auWikepedia |
Standardization Method Acid Base TitrationBy: Alleta Fae S. LiwagStandardization is the process of determining the exact concentration (molarity) of a solution. Titration is one type of analytical procedure often used in standardization. In a titration, an exact volume of one substance is reacted with a known amount of another substance. The point at which the reaction is complete in a titration is referred to as the endpoint. A chemical substance known as an indicator is used to indicate (signal) the endpoint. How to Standardize? Example 1
A 0.128 g sample of KHP (HKC8H4O4) required 28.54 mL of NaOH solution to reach a phenolphthalein endpoint. Calculate the molarity of the NaOH. HKC8H4O4 + NaOH -----> NaKC8H4O4 + H2O (0.128 g KHP)(1 mol / 204.23 g KHP ) = 6.267 x 10-4 mol KHP (6.267 x 10-4 mol KHP)(1 mol NaOH / 1 mol KHP) = 6.267 x 10-4 mol NaOH 6.267 x 10-4 mol NaOH / 0.02854 L NaOH = 0.0220 M NaOH
HCl + NaOH -----> NaCl + H2O (0.02372 L NaOH)(0.0220 mol NaOH / 1 L NaOH) = 5.218 x 10-4 mol NaOH (5.218 x 10-4 mol NaOH)(1 mol HCl / 1 mol NaOH) = 5.218 x 10-4 mol HCl 5.218 x 10-4 mol HCl / 0.02000 L HCl = 0.0261 M HCl REFERENCE: http://www.chem.latech.edu/~deddy/chem104/104Standard.htm |
Tirtation an Acid-Base Neutralization
By: Mark Anthony BasilioAcids and Bases - Neutralizing a Base with an Acid Ca(OH)2 is a strong base and will dissociate completely in water to Ca2+ and OH-. For every mole of Ca(OH)2 there will be two moles of OH-. The concentration of Ca(OH)2 is 0.01 M so [OH-] will be 0.02 M.To the solution will be neutralized when the number of moles of H+ equals the number of moles of OH-. Step 1: Calculate the number of moles of OH-. Molarity = moles/volume moles = Molarity x Volume moles OH- =
0.02 M/100 ml
Step 2: Calculate the Volume of HCl needed Molarity = moles/volume Volume = moles/Molarity A titration is a volumetric technique in which a solution of one reactant (the titrant) is added to a solution of a second reactant (the "analyte") until the equivalence point is reached. The equivalence point is the point at which titrant has been added in exactly the right quantity to react stoichiometrically with the analyte. If either the titrant or analyte is colored, the equivalence point is evident from the disappearance of color as the reactants are consumed. Otherwise, an indicator may be added which has an "endpoint" (changes color) at the equivalence point, or the equivalence point may be determined from a titration curve. The amount of added titrant is determined from itsconcentration and volume: n(mol) = C(mol/L) x V(L) and the amount of titrant can be used in the usual stoichiometric calculation to determine the amount of analyte. The object of a titration is always to add just the amount of titrant needed to consumeexactly the amount of substance being titrated. In the NaOH—CH3COOH reaction [Eq. (1)], the equivalence point occurs when an equal molar amount of NaOH has been added from the graduated cylinder for every mole of CH3COOH originally in the titration flask. That is, at the equivalence point the ratio of the amount of NaOH, added to the amount of CH3COOH consumed must equal the stoichiometric ratio Note that overtitrating [adding more than 23.62 cm3 of KMnO4(aq) would involve an excess (more than 1.272 mmol) of KMnO4. Source: http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Titrations-875.html http://chemistry.about.com/od/workedchemistryproblems/a/neutralizeacid.htm |
Indirect TirtationBy: Rosiel MarianoBack titration (Indirect titration) is
a titration done in reverse;; instead of titrating the original sample,
a known excess of standard reagent is added to the solution, and the
excess is titrated. A back titration is useful if the endpoint of the
reverse titration is easier to identify than the endpoint of the normal
titration, as with precipitation reactions.
Back titrations are also useful if the reaction between the analyte and
the titrant is very slow, or when the analyte is in a non-soluble solid. PROPERTIES OF A BACK TITRATION •React
with the excess volume of reactant which has been left over after completing
reaction with the analyte from the normal titration •The
substance or solution of unknown concentration of excess intermediate reactant
is made to react with known volume and concentration of intermediate reactant
solution in back titration•Throughout
back titration, the reaction can reach the completion quickly as the excess
reactant that react with the NaOH (as example) heated, and is much easier
to measure •Back
titration also an indirect titration procedure •the
proportion consumed in the reaction of back titration being obtained by
difference PURPOSE OF BACK TITRATION Back titration is designed to resolve
some of the problems encountered with forward or direct titration. Possible
reasons for devising back titration technique are: •1: The analyte may be in solid form
•2: The analyte may contain
impurities which may interfere with direct titration. Consider the case of
contaminated chalk. We can filter out the impurities before the excess reactant
is titrated and thus avoid this situation. •3: The analyte reacts slowly with
titrant in direct or forward titration. The reaction with the intermediate
reactant can be speeded up and reaction can be completed say by heating. •4: Weak acid –
weak base reactions can be subjected to back titration for analysis of solution
of unknown concentration. Recall that weak acid-weak weak titration does not
yield a well defined change in pH, which can be detected using an indicator. ADVANTAGES OF BACK TITRATION •useful when trying to
work out the amount of an acid or base in a non-soluble solid. •useful if the endpoint
of the reverse titration is easier to identify than the endpoint of the normal
titration DISADVANTAGES OF BACK TITRATION •Needs
skill and practise for effective results •Instruments
have to be properly calibrated since it will give affected the final result. •Reactivity
of the elements to be titrated should be well researched since this may affect
the end point. •Time
consuming if done manually CALCULATIONS Example 1 150.0
mL of 0.2105 M nitric acid was added in excess to 1.3415 g calcium
carbonate. The excess acid was back titrated with 0.1055 M sodium
hydroxide. It required 75.5 mL of the base to reach the end point.Calculate the percentage (w/w) of calcium carbonate in the sample. 1.EXTRACT INFORMATION •HNO3 V=150.0 mL M=0.2105 M •CACO3
Mass= 1.3415 g •NAOH
M=0.1055 M V=75.5 mL 2. Write balanced equation 2HNO3 + CaCO3 ® Ca(NO3)2 + CO2 + H2O ------ 1
HNO3 + NaOH ® NaNO3 + H2O ------- 2 2 mole of HNO3 react with 1 mole of CaCO3 1 mole of HNO3 react with 1 mole of NaOH 3. Calculate no of mole Initial amount HNO3: No of mole of acid = 0.2105 x 150 = 31.575 mole acid. Excess acid No of mole of excess acid = 0.1055 x 75.5 = 7.965 mmole acid mole of acid reacted with CaCO3 = ( 31.575 – 7.965 ) = 23.61 mole acid 4. Mole ratio 2 mole of HNO3 react with 1 mole of CaCO3 Thus,23.61 mole of HNO3 react with ½(23.61) mole of CaCO3 mole of CaCO3 = ½ x mole acid = ½ x 23.61 = 11.805 mole CaCO3. 5. Find mass
Gram CaCO3 = mole x molar mass = 11.805 x 10-3 x 100 = 1.1805 g. 6.Find percentage % CaCO3 = Ï 100
= Ï 100 = 87.99 % (w/w) REFERENCES http://en.wikipedia.org/wiki/Titration#Back_titration |
Red-ox ReactionBy: Reagan Delos ReyesRedox reactions(oxidation-reduction reactions)- have a number of similarities to acid-base reactions. Fundamentally, redox reactions are a family of reactions that are concerned with the transfer of electrons between species. Like acid-base reactions, redox reactions are a matched set -- you don't have an oxidation reaction without a reduction reaction happening at the same time. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. Each reaction by itself is called a "half-reaction", simply because we need two (2) half-reactions to form a whole reaction. In notating redox reactions, chemists typically write out the electrons explicitly: Cu (s) ----> Cu2+ + 2 e- This half-reaction says that we have solid copper (with no charge) being oxidized (losing electrons) to form a copper ion with a plus 2 charge. Notice that, like the stoichiometrynotation, we have a "balance" between both sides of the reaction. We have one (1) copper atom on both sides, and the charges balance as well. The symbol "e-" represents a free electron with a negative charge that can now go out and reduce some other species, such as in the half-reaction: 2 Ag+ (aq) + 2 e- ------> 2 Ag (s) Here, two silver ions (silver with a positive charge) are being reduced through the addition of two (2) electrons to form solid silver. The abbreviations "aq" and "s" mean aqueous and solid, respectively. We can now combine the two (2) half-reactions to form a redox equation: As a summary, here are the steps to follow to balance a redox equation in acidic medium (add the starred step in a basic medium): 1. Divide the equation into an oxidation half-reaction and a reduction half-reaction 2. Balance these o Balance the elements other than H and O o Balance the O by adding H2O o Balance the H by adding H+ o Balance the charge by adding e- 3. Multiply each half-reaction by an integer such that the number of e- lost in one equals the number gained in the other 4. Combine the half-reactions and cancel 5. **Add OH- to each side until all H+ is gone and then cancel again** |
Back Tirtation By: Kathleen Caralde
What is indirect titration? This method employs a preliminary reaction in which the analyte is replaced by an equivalent amount of another substance which is then determined by titration. The titrant and analyte do not react with each other, but are related through the other substance. e.g. Determination of iron Fe+3+2I-2Fe+2+12 (analyte) I2+2S2O3-2I+S4O6-2 (titrant) It is generally a two-stage analytical technique: a. Reactant A of unknown concentration is reacted with excess reactant B of known concentration. b. A titration is then performed to determine the amount of reactant B in excess. Or scientifically speaking.. Preliminary reaction: aA + rR à pP Titration reaction: pP + tT à fF Imdirect titrations are used when: * one of the reactants is volatile, for example ammonia. * an acid or a base is an insoluble salt, for example calcium carbonate * a particular reaction
is too slow * direct titration would involve a weak acid - weak base titration
(the end-point of this type of direct titration is very difficult to observe)
Example : Indirect Titration to Determine the Concentration of a Volatile Substance A student was asked to determine the concentration of ammonia, a volatile substance, in a commercially available cloudy ammonia solution used for cleaning. First the student pipetted 25.00mL of thecloudy ammonia solution into a 250.0mL conical flask. 50.00mL of 0.100M HCl(aq) was immediately added to the conical flask which reacted with the ammonia in solution. The excess (unreacted) HCl was then titrated with 0.050M Na2CO3(aq). 21.50mL of Na2CO3(aq) was required. Calculate the concentration of the ammonia in the cloudy ammonia solution. Step 1: Determine the amount of HCl in excess from the titration results a. Write the equation for the titration: 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l) acid + carbonate → salt + carbon dioxide + water b. Calculate the moles, n, of Na2CO3(aq) that reacted in the titration: n = M x V c. M = 0.050 molL-1 d. V = 21.50mL = 21.50 x 10-3L n(Na2CO3(aq)) = 0.050 mol/L x 21.50 x 10-3 L = 1.075 x 10-3 mol e. Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration. From the balanced chemical equation, 1 mole Na2CO3 react with 2 moles of HCl So, 1.075 x 10-3 mol Na2CO3 reacted with 2 x 1.075 x 10-3 moles HCl n(HCltitrated) = 1.075 x 10-3 mol HCL x 2 mol Na2CO3 1mol HCl = 2.150 x 10-3 mol f. The amount of HCl
that was added to the cloudy ammonia solution in excess was 2.150 x 10-3 mol Step 2: Determine the amount of ammonia in the cloudy ammonia solution a. Calculate the total moles of HCl originally added to the diluted cloudy ammonia solution: n(HCltotal added) = M x V M = 0.100 molL V = 50.00mL = 50.00 x 10-3L n(HCltotal added) = 0.100 molL x 50.00 x 10-3 L = 5.00 x 10-3 mol b. Calculate the moles of HCl that reacted with the ammonia in the diluted cloudy ammonia solution n(HCltitrated) + n(HClreacted with ammonia) = n(HCltotal added) n(HCltotal added) = 5.00 x 10-3 mol n(HCltitrated) = 2.150 x 10-3 mol 2.150 x 10-3 + n(HClreacted with ammonia) = 5.00 x 10-3 n(HClreacted with ammonia) = 5.00 x 10-3 - 2.150 x 10-3 = 2.85 x 10-3 mol c. Write the balanced chemical equation for the reaction between ammonia in the cloudy ammonia solution and the HCl(aq). NH3(aq) + HCl(aq) → NH4Cl(aq) d. From the balanced chemical equation, calculate the moles of NH3 that reacted with HCl. From the equation, 1 mol HCl reacts with 1 mol NH3 So, 2.85 x 10-3 mol HCl had reacted with 2.85 x 10-3 mol NH3 in the cloudy ammonia solution. e. Calculate the ammonia concentration in the cloudy ammonia solution. M = n ÷ V n = 2.85 x 10-3 mol (moles of NH3 that reacted with HCl) V = 25.00mL = 25.00 x 10-3L (volume of ammonia solution that reacted with HCl) M = 2.85 x 10-3 ÷ 25.00 x 10-3 = 0.114 M f. The concentration of ammonia in the cloudy ammonia solution was 0.114M. |
Half Reaction(Review)
By: Cristel Diane Dela Cruz
Half-reaction is the part of an overall reaction that represents, separately, either an oxidation or a reduction. - a half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction. -a half reaction does not occur by itself, at least two such reactions must be coupled so that the electron released by one reactant is accepted by another in order to complete the reaction. -it is a method for balancing oxidation/reduction reactions that occur in aqueous solution -one of the half reaction will illustrate reduction that means the electrons will show up in the reactant side A + e- -> A- -one of the half reaction will illustrate oxydation that means the electrons will show up in the product side B-> B+ + e- Remember : The number of electrons lost by the oxidized species must be equal to the number of electrons gained by the reduced species. Example : When a nickel strip {Ni (s)} is placed in an aqueous solution of copper(II) sulfate {Cu2+, SO42-}, an immediate reaction occurs. Copper metal begins to deposit on the strip. The only source for metallic copper in this system is the copper(II) ions in solution. What is happening to the copper(II) ions to cause them to change into elemental copper? Let's represent this fascinating observation with a chemical equation: We have just written a half-reaction! In this equation, the copper(II) ion is being reduced When we balance a half-reaction, we first balance the mass of the participating species (atoms, ions, or molecules) and then the charge. In this case, the mass is balanced by adding a copper (atom or ion) to each side. To balance the charge, electrons are added. Notice the addition of 2 electrons to left side of the above equation. These electrons are necessary to reduce a copper(II) ion to metallic copper. If copper(II) ion is being reduced, what is being oxidized? Another way to ask this question is 'where are those electrons coming from?'. Nickel, of course! What is the half-reaction for the oxidation of metallic nickel? Nickel must release 2 electrons to form the nickel(II) ion as shown in the following equation (remember that we balance the mass and then the charge): REFERENCES: http://www.youtube.com/watch?v=zKT7QZbto34 |
SECOND GRADING |
Unfollowing the Octet Rule of 8By: Cristel Imbag An atom with less than eight electrons in its valence shell. An atom with less than eight total bonding and nonbonding electrons in a Lewis structure, for example, B in BH3 has an incomplete octet. There are certain atoms of certain elements that can exist in stable compounds forming bonds with less than eight valence electrons. When this occurs, the atom of the element within the molecule is said to contain an incomplete octet. The common examples of such elements are hydrogen (stable with only 2 valence electrons), beryllium (stable with only 4 valence electrons) and boron and aluminum (stable with only 6 valence electrons). For hydrogen 2 valence electrons give it a noble gas structure (like He) so this is much like the octet rule for everything else below period 1. But covalent compounds in groups 2 and 3 can form stable compounds in which the valence electrons are not in the noble gas structure. However, for these compounds you will find that they do form compounds that minimize formal charge.
Example: Good examples of the first type of exception
are provided by BeCl2 and BCl3. Beryllium dichloride,
BeCl2, is a covalent rather than an ionic substance. Solid BeCl2
has a relatively Complex structure at room temperature, but when
it is heated to 750°C, a vapor which
consists of separate BeCl2 molecules is obtained. Since Cl atoms do
not readily form multiple bonds, we expect the Be atom to be joined to each Cl
atom by a single bond. The structure is Instead of an octet the valence shell of Be contains only two electron pairs. Similar arguments can apply to boron trichloride, BCl3, which is a stable gas at room temperature. We are forced to write its structure as
Expanded octet A hypervalent molecule (the phenomenon is sometimes colloquially known as expanded octet) is a molecule that contains one or more main group elements formally bearing more than eight electrons in their valence shells. Phosphorus pentachloride (PCl5), sulfur hexafluoride (SF6), chlorine trifluoride (ClF3), and the triiodide (I3−) ion are examples of hypervalent molecules. Examples : SF4 1. This compound is covalent. 2. Determine the total number of valence electrons available: One sulfur has 6 valence electrons 3. Organize the atoms so there is a central atom (usually the least electronegative) surrounded by ligand (outer) atoms. Hydrogen is never the central atom.
I put the fluorines like I did because I knew I needed an open space for the unbonded pair on the sulfur. If you put the 4 fluorines around the S like in CH4, that's OK. It gets tiresome making all this little graphic files, but hey, I knew that when I took on this project. OK, enough complaining, back to work. 4. Determine a provisional electron distribution by arranging the electron pairs (E.P.) until all available pairs have been distributed:
5. The formal charge (F) on the central atom is zero. The right-hand structure in step 4 is the correct answer. Odd molecule Term invented by Gilbert N. Lewis in 1916 for a molecule containing an odd number of electrons. Taking the p-shell elements, such molecules are rare; they are usually colored and paramagnetic, that is, attracted by a magnet. Odd molecules are 'radicals.' A fine example is nitric oxide, q.v.; nitrogen dioxide is another; chlorine dioxide is also an example, being a reddish-yellow gas. They are all fairly reactive. When including d-shell elements, i.e., the transition metals, the concept mostly doesn't apply, and this 'odd' state is not so unusual. Example: 1. Nitrogen dioxide (NO2) This molecule has a total of 17
electrons to place - five from the nitrogen and 12 from the oxygens. I will go
immediately to the final structure: Notice that I show both resonance structures. Since there are three electron domains, this is a trigonal planar arrangement, but it is signified AX2e, to signal the single electron domain, also called a half-filled orbital. The bond angles are not 120°, since the repulsive power of the single electron is less tha if there were two. So, the O-N-O bond angle moves outward to 134.3°. Adding another electron to make NO2¯ (which creates a full non-bonding electron pair, changes the O-N-O angle to 115.4° and removing an electron (to make NO2+) creates an O-N-O bond angle of 180°. |
Geometry MoleculesBy: Alleta Fae S. Liwag
Geometry of Molecules Chart
Steps Used to Find the Shape of the MoleculeTo sum up there are four simple steps to apply the VSEPR theory. 1. Draw the Lewis Structure. 2. Count the number of electron groups and identify them as bond pairs of electron groups or lone pairs of electrons. Remember electron groups include not only bonds, but also lone pairs! 3. Name the electron-group geometry. (State whether it is linear, trigonal-planar, tetrahedral, trigonal-bipyramidal, or octahedral.) 4. Looking at the positions of other atomic nuclei around the central determine the molecular geometry. (See how many lone pairs there are.) (Use the chart as a guide) Reference: http://chemwiki.ucdavis.edu/Wikitexts/UCD_Chem_2A/ChemWiki_Module_Topics/Chemical_Bonding/Geometry_of_Molecules |
orbital hybridizationBy: cristel diane dela cruz In chemistry hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for the qualitative description of atomic bonding properties. Hybridised orbitals are very useful in the explanation of the shape of molecular orbitals for molecules. It is an integral part of valence bond theory. The hybrids are named based on the atomic orbitals that are involved in the hybridization, and the geometries of the hybrids are also reflective of those of the atomic-orbital contributors Chemist Linus Pauling first developed the hybridization theory in order to explain the structure of molecules. This concept was developed for such simple chemical systems, but the approach was later applied more widely, and today it is considered an effective heuristic for rationalizing the structures of organic compounds. EXAMPLES : Methane, the three 2p orbitals of the carbon atom are combined with its 2s orbital to form four new orbitals called "sp3" hybrid orbitals. The name is simply a tally of all the orbitals that were blended together to form these new hybrid orbitals. Four hybrid orbitals were required since there are four atoms attached to the central carbon atom. These new orbitals will have an energy slightly above the 2s orbital and below the 2p orbitals as shown in the following illustration. Notice that no change occurred with the 1s orbital. Ammonia, the three 2p orbitals of the nitrogen atom are combined with the 2s orbital to form four sp3 hybrid orbitals. The non-bonded electron pair will occupy a hybrid orbital. Again we need a hybrid orbital for each atom and pair of non-bonding electrons. Ammonia has three hydrogen atoms and one non-bonded pair of electrons when we draw the electron-dot formula. In order to determine the hybridization of an atom, you must first draw the electron-dot formula. |
homogenous equilibrium
By: Mark Anthony Basilio ------The equilibrium between different chemical species present in the same or different phases is called chemical equilibrium. There are two types of chemical equilibrium. The equilibrium reactions in which all the reactants and the products are in the same phase are called homogeneous equilibrium reactions.
C2H5OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l) N2(g) + 3H2(g) 2NH3(g) 2SO2(g) + O2(g) 2SO3(g) The equilibrium system, ethanol-acetic acid-ethyl acetate-water, is well known. The equation for this equilibrium is: C2H5OH + CH3COOH <-----> CH3COOC2H5 + H2O Equilibrium is established very slowly at room temperature when these four substances are mixed. The rate of reactions (both forward and reverse) can be accelerated by using a catalyst so that the equilibrium can be more rapidly established. 3 M HCl is used in this experiment as a catalyst, and the equilibrium at room temperature should be established in a few days. This means that in order to successfully complete this experiment, a series of combinations of reactants and/or products must be prepared a few days before the actual experiment. The equilibrium constant is then calculated by the equation: Using aA<----->bB Kc=[B]a/[A]b We derive this Kp=PbB/PaB to get Kp=Kc(RT)N Kc = [pure ethyl acetate] [water] / [ethanol] [glacial acetic acid] The four quantities in the above equation may be replaced by the number of moles of the four substances involved since the volume is constant in each mixture. Using aA<----->bB Kc=[B]a/[A]b We derive this formula Kp=PbB/PaB to get Kp=Kc(RT)N |
Gas Equilibrium Constants, Kc And KpKc and Kp are the equilibrium constants of gaseous mixtures. -The Equilibrium
Constant, Kc, relates to a chemical reaction at equilibrium. It can
be calculated if the equilibrium concentration of each reactant and product in
a reaction at equilibrium is known.
Kc=molar concentration
of products/molar concentration of reactants
with each concentration raised to the power of the corresponding stoichiometric coefficient. -If we express the
equilibrium constant in terms of partial pressures instead of molar
concentrations, the result is still a constant. The equilibrium constant in
terms of partial pressures is denoted as Kp. THIS ARE SOME WAYS TO WRITE THE GAS EQUILIBRIUM CONSTANT1. In equilibrium equations, even though the arrows point both ways () we usually associate the right as reactants and the left as products. 2. The products are on the TOP of the fraction (the numerator). 3. The reactants are on the BOTTOM of the fraction (the denominator). 4. The concentrations of the products and reactants are always raised to the power of their coefficient in the balanced chemical equation. 5. If any of the reactants or products are solids or liquids, their concentrations are equal to one because they are pure substances. SAMPLE PROBLEMS WITH ANSWERS A 1.) NH4SH (s) NH3 (g) + H2S (g) Kc = but since NH4SH is a solid, we get:
Kc = As for Kp, it is the same as Kc, but instead of brackets [ ], Kp uses parentheses ( ): Kp =
Kp = Kp = 2.) H2 (g) + I2 (g) 2HI (g) Kc = REFERENCEKp = http://chemwiki.ucdavis.edu/Physical_Chemistry/Chemical_Equilibrium/The_Equilibrium_Constant/Calculating_An_Equilibrium_Concentration_From_An_Equilibrium_Constant/Writing_Equilibrium_Constant_Expressions_Involving_Gases/Gas_Equilibrium_Constants,_Kc_And_Kp#Problems |
lewis acid and bases
By: cristel diane dela cruz
The Lewis definitions of acids and bases are even more inclusive than the Bronsted definitions. The Lewis definitions are that an acid is an electron-pair acceptor and a base is an electron-pair donor.
The term Lewis acid refers to a definition of acid published by Gilbert N. Lewis in 1923, specifically: An acid substance is one which can employ an electron lone pair from another molecule in completing the stable group of one of its own atoms. The term Lewis acid refers to a definition of acid published by Gilbert N. Lewis in 1923, specifically: An acid substance is one which can employ an electron lone pair from another molecule in completing the stable group of one of its own atoms. The modern-day definition of Lewis acid, as given by IUPAC is a molecular entity (and the corresponding chemical species) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base.[2] This definition is both more general and more specific—the electron pair need not be a lone pair (it could be the pair of electrons in a π bond, for example), but the reaction should give an adduct (and not just be a displacement reaction). A Lewis base, then, is any species that donates a pair of electrons to a Lewis acid to form a Lewis adduct.
The Lewis acid-base theory can also be used to explain why nonmetal oxides such as CO2 dissolve in water to form acids, such as carbonic acid H2CO3. CO2(g) + H2O(l) H2CO3(aq) In the course of this reaction, the water molecule acts as an electron-pair donor, or Lewis base. The electron-pair acceptor is the carbon atom in CO2. When the carbon atom picks up a pair of electrons from the water molecule, it no longer needs to form double bonds with both of the other oxygen atoms as shown in the figure below One of the oxygen atoms in the intermediate formed when water is added to CO2 carries a positive charge; another carries a negative charge. After an H+ ion has been transferred from one of these oxygen atoms to the other, all of the oxygen atoms in the compound are electrically neutral. The net result of the reaction between CO2 and water is therefore carbonic acid, H2CO3. |
bronsted-lowry acid-base theory
By: zhenna Marriz Aypa In this section we will consider the Brønsted-Lowry concept. This concept focuses on what an acid or base does.In chemistry, the Brønsted–Lowry theory is an acid-base theory, proposed independently by Johannes Nicolaus Brønsted and Thomas Martin Lowry in 1923. The Bronsted-Lowery concept of acids and bases is that acid-base reactions can be seen as proton-transfer reactions. This results in acids and bases being able to be defined in terms of this proton (H+) transfer. In this system,Brønsted acids and Brønsted bases are defined, by which an acid is a molecule or ion that is able to lose, or "donate," a hydrogen cation (proton, H+), and a base is a species with the ability to gain, or "accept," a hydrogen cation (proton). An Acid is a Proton Donor and a Base is a Proton Acceptor So what does this mean? For a reaction to be in equilibrium a transfer of electrons needs to occur. The acid will give an electron away and the base will receive the electron. Acids and Bases that work together in this fashion are called a conjugate pair made up of conjugate acids and conjugate bases. So what does this look like? (A stands for an Acidic compound and Z stands for a Basic compound) HA + Z ↔ A- + HZ+ A Donates H to form HZ+. Z Accepts H from A which forms HZ+ A- becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibrium HZ+ becomes
a conjugate acid of Z and in the reverse reaction it donates a H to
A- recreating Z in order to remain in equilibrium Properties of acids and bases It follows that, if a compound is to behave as an acid, donating a proton, there must be a base to accept the proton. So the Brønsted–Lowry concept can be defined by the reaction: acid + base conjugate base + conjugate acid. The conjugate base is
the ion or molecule remaining after the acid has lost a proton, and the conjugate acid is
the species created when the base accepts the proton. The reaction can proceed
in either forward or backward direction; in each case, the acid donates a
proton to the base. Example:
What happens in this reaction in aqueous solution is that a proton is transferred from H3O+ to NH3. This results in H3O+ losing a (H+), resulting in H2O. The NH3 gains the transferred proton, resulting in NH4+. We call H3O+ the proton donor, or acid. We call NH3 the proton acceptor, or base
The Bronsted-Lowery concept defines something as either an acid or base
depending on its function in the acid-base (proton transfer) reaction. Some
things can act as either an acid or a base. These are called amphiprotic species, they can either lose or gain a proton,
depending on the other reactant. An example of an amphiprotic species would be:
In the Bronsted-Lowery concept we
have found that:
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strengths Of acids and basesBy: kathleen caralde What Makes a Strong Acid or Strong Base? Strong electrolytes are completely dissociated into ions in water. The acid or base molecule does not exist in aqueous solution, only ions. Weak electrolytes are incompletely dissociated. Strong Acids Strong acids are strong electrolytes that, for practical purposes, are assumed to ionize completely in water. Strong acids completely dissociate in water, forming H+ and an anion. There are six strong acids. The others are considered to be weak acids. Most of them are inorganic acids. You should commit the strong acids to memory: · HCl - hydrochloric acid · HNO3 - nitric acid · H2SO4 - sulfuric acid NOTE: this is a diprotic; we show only the first stage of ionization here. · HBr - hydrobromic acid · HI - hydroiodic acid · HClO4 - perchloric acid Weak Acids A weak acid only partially dissociates in water to give H+ and the anion. Examples of weak acids include hydrofluoric acid, HF, and acetic acid, CH3COOH. Weak acids include: · Molecules that contain an ionizable proton. A molecule wih a formula starting with H usually is an acid. · Organic acids containing one or more carboxyl group, -COOH. The H is ionizable. · Anions with an ionizable proton. (e.g., HSO4- → H+ + SO42-) · Cations · transition metal cations · heavy metal cations with high charge · NH4+ dissociates into NH3 + H+ Strong Bases Strong bases are all strong electrolytes that ionizes completely in water. It dissociate 100% into the cation and OH- (hydroxide ion). The hydroxides of the Group I and Group II metals usually are considered to be strong bases. · LiOH - lithium hydroxide CsOH - cesium hydroxide · NaOH - sodium hydroxide *Sr(OH)2 - strontium hydroxide · KOH - potassium hydroxide Ba(OH)2 - barium hydroxide · RbOH - rubidium hydroxide *Ca(OH)2 - calcium hydroxide
* These bases completely dissociate in solutions of 0.01 M or less. The other bases make solutions of 1.0 M and are 100% dissociated at that concentration. There are other strong bases than those listed, but they are not often encountered. Weak Bases Weak bases, like weak acids, are weak electrolytes. Examples of weak bases include ammonia, NH3 (note: ammonia does not ionizelike an acid because it does not split up to form ions the way, say, HCL does), and diethylamine, (CH3CH2)2NH. · Most weak bases are anions of weak acids. · Weak bases do not furnish OH- ions by dissociation. Instead, they react with water to generate OH- ions.
TABLE1 lists some important conjugate acid-base pairs, in order of their relative strengths. Conjugate acid-base have the following properties:
1. If an acid is strong, its conjugate acid-base has no measurable strength. Thus, the ion, which is the conjugate base of the strong acid HCl, is an extremely weak base. 2. is the strongest acid that can exist in aqueous solution. Acids stronger than react with water to produce and their conjugate bases. Thus, HCl, which is a stronger acid than , reacts with water completely to form and : HCl(aq) + (aq) +
Acids weaker than react with water to a much smaller extent, producing and their conjugate bases. For example, the following equilibrium lies primarily to the left: HF(aq) + (aq) +
3. The ion is the strongest base that can exist in aqueous solution. Bases stronger than react with water to produce and their conjugate acids. For example, the oxide ion () is a stronger base than , so it reacts with water completely as follows: (aq) + For this reason the oxide ion does not exist in aqueous solutions. EXAMPLE: Calculate the pH of (a) a M HCl olution and (b) a 0.020 M solution. STRATEGY: Keep in mind that HCl is a strong acid and is a strong base. Thus, these species are completely ionized and no HCl and will be left in solution. |
FIRST GRADING |
The Mole Concept by: Zhenna Marriz Aypa In chemistry the mole
is a fundamental unit in the Système International d'Unités, the SI system, and
it is used to measure the amount of substance. This quantity is sometimes
referred to as the chemical entity, which can be an atom, molecule, formula unit, electron or proton. In Latin mole means a "massive heap" of material. It
is convenient to think of a chemical mole as such.
1 mole
C à 6.02 x 1023
atoms 1 mole NaCl à 6.02 x 1023 molecules
Atoms and molecules are incredibly small and even a tiny chemical sample contains an unimaginable number of them. Therefore, counting the number of atoms or molecules in a sample is impossible. The multiple interpretations of the mole allow us to bridge the gap between the submicroscopic world of atoms and molecules and the macroscopic world that we can observe To determine the chemical amount of a sample, we use the substance's molar mass, the mass per mole of particles. General Plan for Converting Mass, Amount, and Numbers of ParticlesSample Problem 1:
How many moles of Magnesium are 3.01 x 1022 atoms of magnesium? Solution: Analyze: Given: 3.01 x 1022 atoms of Mg Mole of Mg= ( ?) Compute: 3.01 x 1022 atoms of Mg x 1 mol of Mg 6.02 x 1023 atoms of Mg = 0. 500 Mol Mg Sample Problem No. 2 How many moles are there in 29.0 grams of Sodium Chloride? Analyze: 29.0 g NaCl à mole(?) (MM): MM = 22.99 + 35.45 = 58.44 g NaCl Compute: 29.0 g Nacl x 1 mol NaCl 58.44 g NaCl = 0.496 mol NaCl Sample Problem No. 3 How many grams are in 7.20 mol of Dinitrogen Trioxide? Given: 7.20 mol N2O3 à g(?) (MM): N2(14.01 x 2) + O3(16x 3) = 76.02 mol of N2O3 Compute: 7.20 mol of N2O3 x 1 mol 76.02 mol of N2O3 = 0.0947 mol N2O3
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Stoichiometryby: Jarlyn Liwag→ " Stoichiometry " was derived from: "Stoicheion" meaning element and "Metron" meaning measure. · deals with the relative quantities of Reactants and Products in chemical reactions. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of whole numbers.
· used to find the right amount of reactants to use in a chemical reaction. Example : *Consider the complete combustion of natural gas Methane ( CH4 ) reacts with Oxygen gas ( O2 ) to produce CO2 and water ( H2O ). CH4 + 2O2 → CO2 + 2H2O *Hydrogen gas (H2)
reacts with Nitrogen gas (N2) to give ammonia (NH3) 3H2+ N 2 →2NH3
· often used to balance chemical equations .
*For example, the two diatomic
gases, hydrogen and oxygen, can combine to form
a liquid, water (H2O). 2H2 + O2 → 2H2O - describes the 2:1:2 ratio of hydrogen, oxygen, and water molecules in the above equation.
· is simply the study of the relationships or ratios between two or more substances undergoing a physical or chemical change (chemical reaction). → In Stoichiometry, we have different PROBLEMS.
· Mole-Mole Problems. · Mass-Mole Problems. · Volume-Volume Problems.
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Limiting and Excess Reagent
By: Alleta Fae S. Liwag · Limiting Reactant - The reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed. · The limiting reagent is the reactant that is completely used up during the chemical reaction. · Excess Reactant - The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed. The excess reactant remains because there is nothing with which it can react.
No matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made. Likewise with chemistry, if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up. Example Limiting Reactant Calculation: A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? First, we need to create a balanced equation for the reaction:
Next we can use stoichiometry to calculate how much product is produced by each reactant. NOTE: It does not matter which product is chosen, but the same product must be used for both reactants so that the amounts can be compared.
The reactant that produces the lesser amount of product in this case is the oxygen, which is thus the "limiting reactant." Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant (ammonia) actually did react with the limiting reactant (oxygen).
We're not finished yet though. 1.70 g is the amount of ammonia that reacted, not what is left over. To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample.
Reference: http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm |
Mass - Mass Stoichiometry by: Cristel Diane Dela Cruz A mass to mass problem is one in which the mass of a reactant or product is given. You are then asked to calculate the mass of another reactant required or the mass of another product formed. There are four steps involved in solving these problems: 1. Make sure you are working with a properly balanced equation. 2. Convert grams of the substance given in the problem to moles. 3.
Construct
two ratios - one from the problem and one from the equation and set them equal.
Solve for "x," which is usually found in the ratio from the problem. 4. Convert moles of the substance just solved for into grams.
Example : Follow the steps in the diagram Under severe conditions; high temperature, pressure and concentration; hydrogen gas and nitrogen gas can be coerced into forming ammonia, NH3, Also a gas at these temperatures. What mass of N2 would be required if mixed with excess hydrogen gas (meaning more than enough - the reaction will cease when the nitrogen runs out) to produce 850g ammonia? What mass of H2 would be used?
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Gas Stoichiometry by: Cristel Imbag Stoichiometry is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of whole numbers. For example, in a reaction that forms ammonia (NH3), exactly one molecule of nitrogen (N2) reacts with three molecules of hydrogen (H2) to produce two molecules of NH3: N2 + 3H2 → 2NH3
Gas stoichiometry deals with reactions involving gases, where the gases are at a known temperature, pressure, and volume, and can be assumed to be ideal gases. For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, due to the existence of isotopes, molar masses are used instead when calculating the mass ratio. Gas stoichiometry is the quantitative relationship (ratio) between reactants and products in a chemical reaction with reactions that produce gases. Gas stoichiometry applies when the gases produced are assumed to be ideal, and the temperature, pressure, and volume of the gases are all known. The ideal gas law is used for these calculations. Often, but not always, the standard temperature and pressure (STP) are taken as 0 °C and 1 bar and used as the conditions for gas stoichiometric calculations. Gas stoichiometry calculations solve for the unknown volume or mass of a gaseous product or reactant. For example, if we wanted to calculate the volume of gaseous NO2 produced from the combustion of 100 g of NH3, by the reaction: 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (l) we would carry out the following calculations:
There is a 1:1 molar ratio of NH3 to NO2 in the above balanced combustion reaction, so 5.871 mol of NO2 will be formed. We will employ the ideal gas law to solve for the volume at 0 °C (273.15 K) and 1 atmosphere using the gas law constant of R = 0.08206 L · atm · K−1 · mol−1
Gas stoichiometry often involves having to know the molar mass of a gas, given the density of that gas. The ideal gas law can be re-arranged to obtain a relation between the density and the molar mass of an ideal gas: and and thus:
Sample Problems: 1) A quantity of gas at a temperature of 18oC has a volume of 124.5 cm3 at a pressure of 97.8 KPa. How many moles of gas are there? PV=nRT n= [
mol K ][97.8
KPa ][0.1245dm3]
= 0.005 mol gas 2) At
a pressure of 105.2 KPa, 1.42 moles of a gas occupy 1.74 dm3.
What is the T = [
mol K ][105.2 KPa][1.74dm3] =
15.51K 3) A
gas at 1.24oC occupies a 3.51 dm3 container. If
there are 3.34 moles of the gas, P= [3.34 mol][8.31 dm3 KPA][1.24oC
+ 273=K] = 2170 KPa 4) If
1.39 g of carbon monoxide is reacted with oxygen, what volume of carbon dioxide 2CO + O2 --> 2CO2 mol CO2
= [1.39g CO][1mol CO][2 mol CO2 ] = 0.05 mol CO2 V= [0.05
mol][8.31 dm3 KPA][12.3oC + 273=K] = 1.09 dm3 5) If
14.4 dm3 of ethane is combusted at 102.7oC and 99.3KPa,
how many grams of 2C2H6 + 7O2 --> 4CO2 + 6H2O
n= [ mol
K ][99.3
KPa
][14.4 dm3] = 0.46 mol C2H6
mass H2O = [0.46 mol C2H6][6 mol H2O
][18 g H2O] = 24.8 g H2O
REFERENCES: Ø http://goldbook.iupac.org/S06025-plain.html Ø IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8. doi:10.1351/goldbook. Entry: "stoichiometric number". Ø IUPAC Compendium of Chemical chemistry/chprbgsst.htm Terminology 2nd Edition (1997) Ø http://www.sciencebugz.com/
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Balancing Redox Equations
by: Kathleen Caralde
Redox (reduction-oxidation) reactions
include all chemical reactions in which atoms have their oxidation
state changed. A reaction in which oxidation and reduction take place
simultaneously is called a redox
reaction. Oxidation number Method OBJECTIVES: In this article you will learn to write a balanced equation for a redox reaction using Oxidation number Method. Identify the oxidation numbers of the elements in the equation. State the characteristics of a redox reaction and identify the oxidizing agent and reducing agent. Before we will get to explanation, very important disclaimer: oxidation numbers don't exist. They were invented to help in charge accounting needed when balancing redox reaction equations, but they don't refer to any real life chemical concept. The general idea behind the oxidation numbers (ON) method for balancing chemical equations is that electrons are transferred between charged atoms. These charges (the positive or negative number that indicates how many electrons an atom has gained, lost, or shared to become stable) - assigned to individual atoms - are called oxidation numbers, just to remind you that they don't reflect real structure of the reagents. How do we use oxidation numbers for balancing? First of all, we have to understand that oxidation means increase of oxidation number, while reduction means decrease of oxidation number. In both cases change of oxidation number is due to electrons lost (oxidation) or gained (reduction). We calculate oxidation numbers for all atoms present in the reaction equation (note that it is not that hard as it sounds, as for most atoms oxidation numbers will not change) and we look for a ratio that makes the number of electrons lost equal to the number of electrons gained. That gives us additional information needed for reaction balancing.
CHEMISTRY IMPORTANT TERMS REDOX REACTION - oxidation and reduction reactions that occurs simultaneously. -Process to characterize by increasing or decreasing oxidation number. - Losing or gaining of electrons
- It is characterize by an increase of oxidation number REDUCTION - Process by which a substance gains e- - Decrease of oxidation number REDUCING AGENT- substance that causes another substance to be reduced. OXIDIZING AGENT- substance that causes another substance to be oxidized. Reference: http://wiki.answers.com/Q/What_is_an_oxidation_number#ixzz216Q2GDYK http://science.widener.edu/~svanbram/chem146/ch20/redox.html http://www.docslide.com/oxidation-reduction-redox-reactions-2/ http://www.scribd.com/doc/39095944/Chapter-7-Oxidation-and-Reduction-Reactions |