I. Equilibrium Constant for the Dissociation of a Weak Acid--Ka
A. Weak Acid description:
- it is an acid which only partially dissociated in water
- the equation for the partial dissociation of the weak acid HB is HB <____> H+ + B-
- The products of the dissociation are hydrogen ion (H+) and an anion (B-), which is the conjugate base.
- How can one identify a weak acid in general?
a. it has an H at the start of the formula
b. it is not one of the strong acids (You should know these from memory: HCl, HBr, HI, HNO3, H2SO4, HClO4).
B. Expression for Ka
- The equilibrium expression for the dissociation of a weak acid follows the general rules for any equilibrium reaction.
HB <____> H+ + B-
Ka = [H+][B-] / [HB]
Example 1: Acetic acid, HC2H3O2, partially dissociates in water. Write the equation for its dissociation and the equilibrium expression for Ka.
NOTE: C2H3O2- (acetate ion) is often written as OAc- when the interest is in the acetate ion as a whole. The formula for acetic acid, thus would be HOAc.
HOAc <____> H+ + OAc-
Ka = [H+] [OAc-] / [HOAc]
NOTE: Very Important!!!! Ka always means dissociation, so H+ is always in the numerator. The anion which results from the dissociation of the weak acid is known as the conjugate base. It accepts the proton (hydrogen ion) and reforms the undissociated molecule.
Example 2: Write the dissociation equation for the dissociation of nitrous acid in water.
HNO2 <____> H+ + NO2-
NO2- is the conjugate base because it accepts the H+ to produce nitrous acid in the reverse reaction.
Write the expression for the Ka.
Ka = [H+][NO2-] / [NHO2]
Weak acids with more than one ionizable H+, H3PO4, for example, dissociate stepwise, one H+ at a time. Each step has a different Ka expression and a different Ka value. The Ka value decreases with each successive step. The steps for the dissociation of H3PO4 are: (all are in aqueous solution)
H3PO4 <____> H+ + H2PO41-
H2PO41- <____> H+ + HPO42-
HPO42- <____> H+ + PO43-
NOTE: the conjugate base for the first step becomes the weak acid of the second step. Now write the equations for the stepwise dissociation of carbonic acid.
H2CO3 <____> H+ + HCO31-
HCO31- <____> H+ + CO32-
The ions and molecules are enclosed in brackets, [ ] , when they are written in the equilibrium expression, just as with other equilibrium reactions. This means that the concentrations are given in moles/liter, or molarity.
The equilibrium concenttration of the undissociated weak acid is equal to the original concentration, (HB) minus the amount of the weak acid that is dissociated.
[HB] = (HB) - [H+]
C. What does Ka tell you?
- the smaller the value of Ka, the weaker the acid is
- when two weak acids are compared, the one with the smaller Ka
a. is the weakest
b. has a smaller [H+]
c. has a higher pH
d. has a larger [HB]
D. Determination of Ka
- Given a known amount of weak acid in a given volume, as well as the pH of the resulting solution.(In other words, the original concentration of the weak acid and the [H+] )
Example 1: A solution of HClO2 is prepared by dissolving 1.369 g of HClO2 in enough water to make 100.0 mL of solution. The pH of the resulting solution is 1.36.
a. First, write the dissociation equation:
HClO2 <____> H+ + ClO2-b. Next, write the Ka expression: Ka = [H+] [ClO2-] / [HClO2]
c. Then calculate Ka by setting up a chart as was done with other equilibrium problems.
species original conc. (M) change equilibrium conc (M)
HClO2 H+ ClO2-
The OC for the HClO2 can be calculated by dividing the mass of the acid by the volume in liters and then by the molar mass, which for HClO2 is 68.46 g/mol.
The change for the acid will have a negative sign and the changes for the H+ and ClO2- will be positive.
The pH is then used to calculate the [H+]
[H+] = inverse log -pH = inverse log(-1.36) = 0.044 M
Since the OC for the H+ was zero, then the change in concentration for the H+ is +0.0437 M. This is also the change for the ClO2-. The change for the HClO2 is a negative 0.0437 M, so the EC would be 0.2000 M - 0.0437 M = 0.01563 M.
species original conc. (M) change equilibriumconc (M) HClO2 0.2000 -0.0437 0.1563 H+ 0 +0.0437 0.0437 ClO2- 0 +0.0437 0.0437
Ka = [H+][ClO2-] / [HClO2] = [0.0437][0.0437] / [0.1563] = 0.0122
Example 2: A one-liter solution of nitrous acid contains 11.76 g of the weak acid. The solution has a pH of 1.955. Calculate Ka for HNO2.
species original conc. (M) change equilibrium conc (M) HNO2 H+ NO2-
Given the original concentration and percent dissociation of the acid.
Example 1: A 0.300 M phosphorous acid solution is 33.3% dissociated in the first-step.
The equation for the first-step dissociation is
H3PO3 <____> H+ + H2PO3-33.3% dissociation means that 33.3% of the phosphorous acid has dissociated or changed into H+ and H2PO3-. The change in the H3PO3 is therefore, (0.300)(0.333) = 0.0999 M. It has a negative sign, and the change for the two ions is a positive 0.100 M each. The remainder of the problem is carried out as any other equilibrium problem.
The Ka expression is: Ka = [H+] [H2PO3-] / [H3PO3]
species original conc. (M) change equilibrium conc (M) H3PO3 0.300 -0.0999 0.200 H+ 0 +0.0999 0.0999 H2PO3- 0 +0.0999 0.0999
Ka = [H+] [H2PO3-] / [H3PO3] = [0.0999][0.0999] / [0.200] = 0.0499
Example 2: The second-step dissociation of a 0.0100 M solution of oxalic acid, H2C2O4, is 7.35% of the HC2O4- ion. Calculate Ka for the second-step dissociation of oxalic acid.
species original conc. (M) change equilibrium conc (M) HC2O4- H+ C2O42-
E. Applications of Ka:
- Determination of the [H+] or pH of a solution of a weak acid when the OC is known.
Example 1: Phenol (C6H5OH) has a Ka = 1.6 x 10-10. Determine the [H+] in a solution prepared by dissolving 0.500 mol of phenol to form 5.00 L of solution.
We don't know any of the EC values, so we let the change for each be "x". The phenol change would be -x, giving the EC for phenol as 0.100 - x. The change for the two ions would be +x for each, with an EC of just x for each.
species original conc. (M) change equilibrium conc (M) C6H5OH 0.100 -x 0.100-x H+ 0 +x x C6H5O- 0 +x x
Substitution into the Ka expression gives:
Ka = [H+] [C6H5O-] / [C6H5OH] = [x] [x] / [0.100-x] = 1.6 x 10-10
Most weak acids have such small dissociation constants that we may assume the amount dissociated is so much smaller than the original concentration that it is negligible when compared to the original concentration. In the preceding equation then we can assume 0.100 - x = 0.100. You have to always write out the assumption and later test whether it is valid.
Assume that x<<0.100
[x][x] / [0.100] = 1.6 x 10-10Substitution of the value of x into the EC of the phenol (0.100 - 0.0000040) shows that the amount of phenol dissociated is negligible compared to the original amount of the phenol. The equilibrium concentration of the phenol is still 0.100 M.
x2 = 1.6 x 10-11
x = 4.0 x 10-6 = [H+]
The assumption must be tested to see if it is valid. The % dissociation must be less than 5%.
% Dissociation = (x / OC of acid) x 100
(0.0000040 / 0.100) x 100 = 0.004 % which is much, much smaller than 5%.
In this case, the assumption was valid, so it was okay to ignore x in the demoninator.
Example 2: sulfurous acid, H2SO3 has a first-step dissociation constant of 1.3 x 10-2.
First write the first-step dissociation equation.
H2SO3 <____> H+ + HSO3-Then calculate the [H+] for a solution with a volume of 5.00 L containing 41.0 g of H2SO3.
species original conc. (M) change equilibrium conc (M) H2SO3 H+ HSO3-
Ka = [H+][HSO3-] / [H2SO3] = 1.3 x 10-2
Notice that the size of Ka is so much larger than that of the previous problem. A good rule to follow is when the size of Ka is greater than or equal to 10-5 then x cannot be ignored in the denominator but if it is smaller than 10-5 it can probably be ignored. The mathematics are certainly simplified when x is ignored.
Determining [H+] in a solution of a weak acid and a conjugate base.
In solving for [H+] in problems where there is a common ion, remember that salts (usually the source of the common ion) in aqueous solutions dissociate completely.
Example 1: Calculate [H+] in a solution with a volume of 1.000 L and which contains 2.70 g of HCN and 2.45 g of NaCN. HCN has a Ka of 4.0 x 10-10.
The common ion is the cyanide ion, CN-, because it is "common" to both the HCN and the NaCN.
The equation for the dissociation of the HCN is as follows:
HCN <____> H+ + CN-And the equation for the dissolution of the NaCN in water is:
NaCN ____> Na+ + CN-The concentration of the HCN is 2.70 g/1.000L x 1 mol/27.0 g = 0.100 M
The NaCN is equal to 2.45 g/1.000L x 1 mol/49.0 g = 0.0500 M
species original conc. (M) change equilibrium conc (M) HCN 0.100 -x 0.100-x H+ 0 +x x CN- 0.0500 +x 0.0500+x
Ka = [H+][CN-] / [HCN-] = [0.100-x][x] / [0.0500] = 4.0 x 10-10
x = 8.0 x 10-10 M = [H+]
Notice that x was left out of the denominator when the values were substituted into the equilibrium expression. The size of the Ka is so small that it seems reasonable to do so.
Notice that the presence of the common ion, cyanide, drives the equilibrium to the left and reduces the amount of hydrogen ion. The HCN at this concentration without the presence of the NaCN would have a [H+] of 6.3 x 10-6 which is considerably larger than with the common ion.
The pH of the HCN-NaCN solution would be 9.1 whereas the HCN alone would show a pH of 5.2. A common ion therefore will serve to lower the [H+] and raise the pH of a solution of a weak acid.
Send questions, comments or suggestions to
Gwen Sibert, at the
Roanoke Valley Governor's School
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