Single Choice

What will be the value of pH of $$0.01 mol\ dm^{-3} CH_3COOH (K_a = 1.74 \times 10^{-5})$$?

A$$3.4$$
Correct Answer
B$$3.6$$
C$$3.9$$
D$$3.0$$

Solution

$$\underset{C(1-\alpha)}{CH_3COOH}+H_2O\longrightarrow \underset{C\alpha}{H_3O}+\underset{C\alpha}{CH_3COO}$$
$$K_a=C\alpha^2\Rightarrow \alpha=\sqrt{\dfrac{Ka}{C}}$$
$$[H_3O^+]=\sqrt[C]{\dfrac{Ka}{c}}=\sqrt{Ka.C}$$
$$=\sqrt{0.01\times 1.74\times 10^{-5}}$$
$$\sqrt{1.74\times 10^{-7}}=4,17\times 10^{-4}$$
$$PH_2-\log(4.17\times 10^{-4})=3.38\simeq 3.4$$

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