Question

What is the pH of 0.2 mole HCl added to 1 litre of the buffer of acetic acid and sodium acetate?

Answer 1

Answer on Question #56522 – Chemistry – Inorganic Chemistry

What is the pH of 0.2 mole HCl added to 1 litre of the buffer of acetic acid and sodium acetate?

Solution:

Processes of dissociation in the solution containing weak $\mathrm{CH}_3\mathrm{COOH}$ acid and its salt:

$$\begin{array}{l} \mathrm{CH_3COOH} \rightleftharpoons \mathrm{CH_3COO^-} + \mathrm{H^+} \\ \mathrm{CH_3COONa} \rightarrow \mathrm{CH_3COO^-} + \mathrm{Na^+} \end{array} \quad \mathrm{Ka} = 1,8*10^{-5}$$

At addition of acid in solution ions of hydrogen communicate in weak acid:

$$\mathrm{H^+} + \mathrm{CH_3COO^-} \rightleftharpoons \mathrm{CH_3COOH}$$

At addition of the basis in solution hydroxide ions communicate in weak electrolyte – water:

$$\mathrm{H^+} + \mathrm{OH^-} \rightleftharpoons \mathrm{H_2O}$$

The constant of dissociation of acid is equal

$$K_a = \frac{[H^+][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}$$ $$\text{Then } [\mathrm{H^+}] = K_a \frac{[\mathrm{CH_3COOH}]}{[\mathrm{CH_3COO^-}]}$$

As degree of dissociation of acetic acid $\alpha << 1$, equilibrium concentration $c[\mathrm{CH_3COOH}]$ is approximately equal to initial concentration with $c(\mathrm{CH_3COOH})$, and equilibrium concentration acetate ions $[\mathrm{CH_3COO^-}]$ is approximately equal to initial concentration of acetate of sodium with $(\mathrm{CH_3COONa})$:

$$[\mathrm{CH_3COOH}] = c(\mathrm{CH_3COOH}), \quad [\mathrm{CH_3COO^-}] = c(\mathrm{CH_3COONa}).$$ $$\text{Then } [\mathrm{H^+}] = K_a \frac{C(acid)}{C(salt)}$$

We will consider that concentration of acetic acid and acetate of sodium are equal in initial buffer solution to $1\ \mathrm{mol/l}$.

After addition of 0,2 mol of HCl concentration of acid in solution will become equal

$$1 + 0,2 = 1,2 \mathrm{~mol/l}$$

Concentration of salt will become equal

$$1 - 0,2 = 0,8 \mathrm{~mol/l}$$

Then pH solution it will be equal

$$\mathrm{pH} = -\lg \mathrm{Ka} - \lg \frac{C(CH_3COOH)}{C(CH_3COONa)} = -\lg (1.8 \times 10^{-5}) - \lg (1,2/0,8) = 4,57$$

**Answer:** pH of a solution is equal 4,57.

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