Question

A weak acid. What is the pH of a 0.1 M solution of acetic acid (pKa = 4.75)?

Answer 1

Answer:

pH of acetic acid solution is 2.88

Explanation:

$pK_{a}=4.75$

or, $-log(K_{a})=4.75$

or, $K_{a}=10^{-4.75}=1.78\times 10^{-5}$

We have to construct an ICE table to determine concentration of $H^{+}$ and corresponding pH. Initial concentration of acetic acid is 0.1 M.

$CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}$

I(M):  0.1                            0                    0

C(M): -x                              +x                 +x

E(M): 0.1-x                         x                    x

So, $\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}=K_{a}$

or, $\frac{x^{2}}{0.1-x}=1.78\times 10^{-5}$

or, $x^{2}+(1.78\times 10^{-5}\times x)-(1.78\times 10^{-6})=0$

So, $x=\frac{-(1.78\times 10^{-5})+\sqrt{(1.78\times 10^{-5})^{2}+(4\times 1\times 1.78\times 10^{-6})}}{2\times 1}$(M)

so, $x=1.33\times 10^{-3}M$

Hence $pH=-log[H^{+}]=-log(1.33\times 10^{-3})=2.88$