Note: in the first three problems, I give the K_{b} of the conjugate base (for example, the acetate ion in Example #1). Often, these problems are given with the K_{a} of the acid and you have to calculate the value of the K_{b}. You do so with this equation:

K_{a}K_{b}= K_{w}

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? K_{b} = 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¯ + H_{2}O ⇌ HAc + OH¯

2) Here is the K_{b} expression for Ac¯:

[HAc] [OH¯] K _{b}=–––––––––– [Ac¯]

3) We can then substitute values into the K_{b} 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 = $\sqrt{\mathrm{(5.65\; x\; 10\xaf10)\; (0.100)}}$

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 K_{b} 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? K_{b} = 2.1 x 10¯^{5}.

**Solution:**

1) Here is the chemical reaction (net ionic) for the hydrolysis of KCN:

CN¯ + H_{2}O ⇌ HCN + OH¯

2) Here is the K_{b} expression for CN¯:

[HCN] [OH¯] K _{b}=––––––––––– [CN¯]

3) We can then substitute values into the K_{b} 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 = $\sqrt{\mathrm{(2.1\; x\; 10\xaf5)\; (0.050)}}$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, C_{6}H_{5}COONa. The K_{b} for C_{6}H_{5}COO¯ (benzoate ion) is 1.55 x 10¯^{10}.

**Solution:**

C

_{6}H_{5}COO¯ + H_{2}O ⇌ C_{6}H_{5}COOH + OH¯

(x) (x) 1.55 x 10¯ ^{10}=––––––– 0.30 − x x = $\sqrt{\mathrm{(1.55\; x\; 10\xaf10)\; (0.30)}}$

x = 6.8 x 10¯

^{6}M <--- the [OH¯]pOH = −log 6.8 x 10¯

^{6}= 5.17pH = 14 − 5.17

pH = 8.83

**Example #4:** Find the pH of a 0.20 M solution of sodium propionate (C_{2}H_{5}COONa), where the K_{a} of propionic acid = 1.34 x 10¯^{5}. (Use the equation K_{w} = K_{a}K_{b} to go from the K_{a} of the acid to the K_{b} of its conjugate base.)

**Solution:**

1) We need to get the K_{b} of the propionate ion first:

1.00 x 10¯ ^{14}K _{b}=–––––––––– = 7.46 x 10¯ ^{10}1.34 x 10¯ ^{5}

2) Now, the solution follows the pattern outlined in the tutorial:

C_{2}H_{5}COO¯ + H_{2}O ⇌ C_{2}H_{5}COOH + OH¯

(x) (x) 7.46 x 10¯ ^{10}=–––––– 0.20 x = $\sqrt{\mathrm{(7.46\; x\; 10\xaf10)\; (0.20)}}$

x = 1.22 x 10¯

^{5}M = [OH¯]pOH = −log 1.22 x 10¯

^{5}= 4.91pH = 14 − 4.91

pH = 9.09

**Example #5:** Given that the K_{a} 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¯ + H_{2}O ⇌ HOCl + OH¯

3) Conjugate pairs have the following relationship:

K_{a}K_{b}= K_{w}(3.5 x 10¯

^{8}) (K_{b}) = 1.0 x 10¯^{14}K

_{b}= 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¯^{4}pOH = −log 2.414 x 10¯

^{4}= 3.62pH = 14 − 3.62 = 10.38

**Example #6:** Calculate the pH of a 0.18 M solution of the weak base NO_{2}¯. The K_{a} of HNO_{2} is 4.5 x 10¯^{4}

**Solution:**

NO_{2}¯ + H_{2}O ⇌ HNO_{2}+ OH¯K

_{b}= ([HNO_{2}] [OH¯]) / [NO_{2}¯]K

_{b}= K_{w}/ K_{a}= 1.00 x 10¯^{14}/ 4.5 x 10¯^{4}= 2.2222 x 10¯^{11}2.2222 x 10¯

^{11}= [(x) (x)] / 0.18x = 0.00000200 M

pOH = −log 0.00000200 = 5.70

pH = 14 − 5.70 = 8.30

**Bonus Example:** Determine the K_{a} of the weak acid HX knowing that a 0.10 M solution of LiX has pH = 8.90.

Solution technique: we will determine the K_{b} of X¯ (since that is what we have data for). Based on the fact that HX is the conjugate base, we will use K_{a}K_{b} = K_{w} to get the K_{a}.

**Solution:**

1) This is the reaction of interest:

X¯ + H_{2}O ⇌ HX + OH¯

2) We need to fill in the right side of the K_{b} 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 K_{b} 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 K_{b} expression. Remember that [X¯] = [OH¯]:

[7.9433 x 10¯ ^{6}] [7.9433 x 10¯^{6}]K _{b}=––––––––––––––––––––––––– 0.10 K

_{b}= 6.3096 x 10¯^{10}

5) The K_{b} of X¯ is related to the K_{a} of HX (what we want to get) by way of K_{w}.

K_{a}K_{b}= K_{w}(K

_{a}) (6.3096 x 10¯^{10}) = 1.0 x 10¯^{14}K

_{a}= 1.5849 x 10¯^{5}Rounded to two significant figures, the final answer is:

K

_{a}= 1.6 x 10¯^{5}