Question

The pH of a 0. 20 M solution of C2H5COONa (Ka C2H5COOH = 1. 34 x 10¯5) is

Answer 1

Answer: The pH is 9.09.

Step-by-step explanation:

This question involves the hydrolysis of a salt into a weak acid and strong base.

Fastest Solution

The fastest solution is found by using the equation:

`\text{pH} = \frac12 \ \text{pK}_w + \frac12 \ \text{pK}_a + \frac12 \log C`

where C represents the concentration of the solution.

We have:

Ka = 1.34 x 10^-5, so pKa = -log(1.34 x 10^-5) = 4.873

Kw = 1.0 x 10^-14, so pKw = -log(1.0 x 10^-14) = 14.00

C = 0.20, so log(C) = log(0.20) = -0.70

Then,

`\text{pH} = \frac12 (4.873) + \frac12 (14.00) + \frac12 (-0.70) = 9.09`

AP Chemistry Level Solution

A solution that uses only fundamental chemistry techniques can be found by considering the chemical reactions involved.

The salt
`\text{C}_2\text{H}_5\text{COONa}` will completely dissolve in solution into its ions:

`\text{C}_2\text{H}_5\text{COONa} \text{ (aq)} \rightarrow \text{C}_2\text{H}_5\text{COO}^- \text{ (aq)} + \text{Na}^+ \text{ (aq)}`

This shows that
`[\text{C}_2\text{H}_5\text{COO}^-] = [\text{C}_2\text{H}_5\text{COONa}] = 0.20 \text{ M}`.

When water is added, an acid-base reaction occurs:

`\text{C}_2\text{H}_5\text{COO}^- \text{ (aq)} + \text{H}_2\text{O} \text{ (l)} \rightleftharpoons \text{C}_2\text{H}_5\text{COOH} \text{ (aq)} + \text{OH}^- \text{ (aq)}`

Using a RICE (Reaction-Initial-Change-Equilibrium) table, we can write the following analysis:

0.20 N/A 0 0 (initial conc.)
- x N/A + x + x (change)
-------------------------------------------------------------------------
(0.20 - x) N/A x x (equilibrium)

Then, we can write the forward base-reaction constant:

`\text{K}_b = \frac{[\text{C}_2\text{H}_5\text{COOH}][\text{OH}^-]}{[\text{C}_2\text{H}_5\text{COO}^-]} = ((x)(x))/(0.20-x) = (x^2)/(0.20-x)`

We have
`\text{K}_a = 1.34 * 10^(-5)` and we also know that
`\text{K}_w = \text{K}_a * \text{K}_b`. It follows that:

`\text{K}_b = \frac{\text{K}_w}{\text{K}_a} = (1.0 * 10^(-14))/(1.34 * 10^(-5)) = 7.46 * 10^(-10)`

Since Kb is very small, we may assume that
`(0.20 - x) \approx 0.20`. Then,

`7.46 * 10^(-10) = (x^2)/(0.20) \quad \rightarrow \quad x \approx \sqrt{0.20(7.46 * 10^(-10))} = 1.22 * 10^(-5)`

Then,

`[\text{OH}^-] = x = 1.22 * 10^(-5)`

`\text{pOH} = -\log(1.22 * 10^(-5)) = 4.91`

Finally,

`\text{pH} = \text{pK}_w - \text{pH} = 14.00 - 4.91 = 9.09`