Question

A buffer is made by dissolving HC2H3O2 and NaC2H3O2 in water.

Write an equation that shows how this buffer neutralizes added acid.
Express your answer as a chemical equation. Identify all of the phases in your answer.
?????

Write an equation that shows how this buffer neutralizes added base.
Express your answer as a chemical equation. Identify all of the phases in your answer.
????
Calculate the pH of this buffer if it is 0.22 M HC2H3O2 and 0.45 M C2H3O2−. The Ka for HC2H3O2 is 1.8×10−5.
Express your answer using two decimal places.
pH = ???

Answer 1

Answer on Question #56388 – Chemistry – General Chemistry

Question:

A buffer is made by dissolving HC2H3O2 and NaC2H3O2 in water.

Write an equation that shows how this buffer neutralizes added acid.

Express your answer as a chemical equation. Identify all of the phases in your answer.

?????

Write an equation that shows how this buffer neutralizes added base.

Express your answer as a chemical equation. Identify all of the phases in your answer.

????

Calculate the pH of this buffer if it is 0.22 M HC2H3O2 and 0.45 M C2H3O2-. The Ka for HC2H3O2 is $1.8 \times 10^{-5}$.

Express your answer using two decimal places.

pH = ???

Solution:

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

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

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 $(\mathrm{CH_3COONa})$:

$$\left[ \mathrm{CH_3COOH} \right] = c \left( \mathrm{CH_3COOH} \right), \quad \left[ \mathrm{CH_3COO^-} \right] = c \left( \mathrm{CH_3COONa} \right).$$

Then $[\mathrm{H}^+] = \mathrm{K}_a \frac{C(CH_3COOH)}{C(CH_3COONa)}$

After logarithming and multiplication on (-1) we will receive

$$\mathrm{pH} = -\lg \mathrm{Ka} - \lg \frac{C(CH_3COOH)}{C(CH_3COONa)} = -\lg (1.8 \times 10^{-5}) - \lg (0.22 / 0.45) = 5.06$$

**Answer:** pH of a buffer it is equal 5.06.

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