Note: in the first three problems, I give the Kb of the conjugate base (for example, the acetate ion in Example #1). Often, these problems are given with the Ka of the acid and you have to calculate the value of the Kb. You do so with this equation:
KaKb = Kw
You will see such a situation starting in the fourth example as well as scattered through the additional problems.
Example #1: What is the pH of a 0.100 M solution of sodium acetate? Kb = 5.65 x 10¯10. (I will use NaAc as shorthand)
Solution:
1) Here is the chemical reaction (net ionic) for the hydrolysis of NaAc:
Ac¯ + H2O ⇌ HAc + OH¯
2) Here is the Kb expression for Ac¯:
[HAc] [OH¯] Kb = –––––––––– [Ac¯]
3) We can then substitute values into the Kb expression in the normal manner:
(x) (x) 5.65 x 10¯10 = –––––––– 0.100 − x
4) If we ignore the minus x (that is, we assume the 5% approximation to be true), we proceed to solve for the hydroxide ion concentration:
(x) (x) 5.65 x 10¯10 ≈ –––––––– 0.100 x =
x = 7.52 x 10¯6 M <--- the [OH¯]
5) We verify the assumption (which is that 0.100 − x ≈ 0.100) by seeing if the following ratio is less than 5%:
[OH¯] ––––––– < 5% [NaAc]o
7.52 x 10¯6 ––––––––– = 0.00752% 0.100 The assumption is valid.
6) We then calculate the pH. Since this is a base calculation, we need to do the pOH first:
pOH = −log 7.52 x 10¯6 = 5.124pH = 14 − 5.124
pH = 8.876
In all the remaining examples we will not repeat the 5% rule calculation. You may verify on your own that in each case, the following is true:
x ––––– < 5% [A¯]o
In other words, you will find that the initial concentrations are large enough and the Kb are small enough that the 5% rule is a valid approximation. If it turns out that the approximation is invalid, then a quadratic equation must be used to solve the problem.
Example #2: What is the pH of a 0.0500 M solution of KCN? Kb = 2.1 x 10¯5.
Solution:
1) Here is the chemical reaction (net ionic) for the hydrolysis of KCN:
CN¯ + H2O ⇌ HCN + OH¯
2) Here is the Kb expression for CN¯:
[HCN] [OH¯] Kb = ––––––––––– [CN¯]
3) We can then substitute values into the Kb expression in the normal manner:
(x) (x) 2.1 x 10¯5 = –––––––– 0.050 − x
4) Ignoring the minus x in the usual manner, we proceed to sove for the hydroxide ion concentration:
x =x = 1.025 x 10¯3 M <--- this is the [OH¯]
5) We then calculate the pH. Since this is a base calculation, we need to do the pOH first:
pOH = −log 1.025 x 10¯3 = 2.99pH = 14 − 2.99
pH = 11.01
Example #3: Find the pH of a 0.30 M solution of sodium benzoate, C6H5COONa. The Kb for C6H5COO¯ (benzoate ion) is 1.55 x 10¯10.
Solution:
C6H5COO¯ + H2O ⇌ C6H5COOH + OH¯
(x) (x) 1.55 x 10¯10 = ––––––– 0.30 − x x =
x = 6.8 x 10¯6 M <--- the [OH¯]
pOH = −log 6.8 x 10¯6 = 5.17
pH = 14 − 5.17
pH = 8.83
Example #4: Find the pH of a 0.20 M solution of sodium propionate (C2H5COONa), where the Ka of propionic acid = 1.34 x 10¯5. (Use the equation Kw = KaKb to go from the Ka of the acid to the Kb of its conjugate base.)
Solution:
1) We need to get the Kb of the propionate ion first:
1.00 x 10¯14 Kb = –––––––––– = 7.46 x 10¯10 1.34 x 10¯5
2) Now, the solution follows the pattern outlined in the tutorial:
C2H5COO¯ + H2O ⇌ C2H5COOH + OH¯
(x) (x) 7.46 x 10¯10 = –––––– 0.20 x =
x = 1.22 x 10¯5 M = [OH¯]
pOH = −log 1.22 x 10¯5 = 4.91
pH = 14 − 4.91
pH = 9.09
Example #5: Given that the Ka for HOCl is 3.5 x 10¯8, calculate the pH of a 0.102 M solution of Ca(OCl)2
Solution:
1) A 0.102 M solution of Ca(OCl)2 is 0.204 M in just OCl¯ ions (which are the conjugate base of the acid HOCl)
2) The chemical reaction of interest is:
OCl¯ + H2O ⇌ HOCl + OH¯
3) Conjugate pairs have the following relationship:
KaKb = Kw(3.5 x 10¯8) (Kb) = 1.0 x 10¯14
Kb = 2.857 x 10¯7
4) For the chemical equilibrium given in step 2, we have:
(x) (x) 2.857 x 10¯7 = –––––– 0.204
5) Cross-multiply, then square root:
x = [OH¯] = 2.414 x 10¯4pOH = −log 2.414 x 10¯4 = 3.62
pH = 14 − 3.62 = 10.38
Example #6: Calculate the pH of a 0.18 M solution of the weak base NO2¯. The Ka of HNO2 is 4.5 x 10¯4
Solution:
NO2¯ + H2O ⇌ HNO2 + OH¯Kb = ([HNO2] [OH¯]) / [NO2¯]
Kb = Kw / Ka = 1.00 x 10¯14 / 4.5 x 10¯4 = 2.2222 x 10¯11
2.2222 x 10¯11 = [(x) (x)] / 0.18
x = 0.00000200 M
pOH = −log 0.00000200 = 5.70
pH = 14 − 5.70 = 8.30
Bonus Example: Determine the Ka of the weak acid HX knowing that a 0.10 M solution of LiX has pH = 8.90.
Solution technique: we will determine the Kb of X¯ (since that is what we have data for). Based on the fact that HX is the conjugate base, we will use KaKb = Kw to get the Ka.
Solution:
1) This is the reaction of interest:
X¯ + H2O ⇌ HX + OH¯
2) We need to fill in the right side of the Kb expression for X¯. The pH is where we start:
pH = 8.90pOH = 14 − 8.90 = 5.10
3) We use the pOH to give us the two values in the numerator of the Kb expression for X¯
[OH¯] = 10¯pOH = 10¯5.10 = 7.9433 x 10¯6 M (keep some guard digits)
4) Now, we can write the Kb expression. Remember that [X¯] = [OH¯]:
[7.9433 x 10¯6] [7.9433 x 10¯6] Kb = ––––––––––––––––––––––––– 0.10 Kb = 6.3096 x 10¯10
5) The Kb of X¯ is related to the Ka of HX (what we want to get) by way of Kw.
KaKb = Kw(Ka) (6.3096 x 10¯10) = 1.0 x 10¯14
Ka = 1.5849 x 10¯5
Rounded to two significant figures, the final answer is:
Ka = 1.6 x 10¯5