Question

0.571 mol of a weak acid, HA, and 11.6 g of NaOH are placed in enough water to produce 1.00 L of solution. The final pH of this solution is 4.14. Calculate the ionization constant, Ka, of HA.

Answer 1

Answer on Question #82833, Chemistry / General Chemistry

0.571 mol of a weak acid, HA, and 11.6 g of NaOH are placed in enough water to produce 1.00 L of solution. The final pH of this solution is 4.14. Calculate the ionization constant, Ka, of HA.

Solution

First of all find the amount of NaOH placed in the solution:

$$v \left(\mathrm{NaOH}\right) = \frac{m}{M} = \frac{11.6}{40} = 0.29 \ (\mathrm{mol})$$

NaOH reacts with weak acid HA which is in excess:

$$\mathrm{NaOH} + \mathrm{HA} \rightarrow \mathrm{NaA} + \mathrm{H_2O}$$

Base and acid react in ratio 1:1, hence the amount of NaA obtained is 0.29 mol.

Find the amount of HA left after the reaction:

$$v \left(\mathrm{HA}\right) = 0.571 - 0.29 = 0.281 \ (\mathrm{mol})$$

After the reaction the solution turns into a buffer with 0.29 mol of NaA and 0.281 mol of HA.

Use the formula for ionization constants of acids in buffers:

$$K_a = \frac{[H^+] \times [A^-]}{[HA]} = \frac{[H^+] \times [NaA]}{[HA]}; \text{ where } [H^+], [HA], [NaA] - \text{concentrations};$$

Find $[H^+]$:

$$\mathrm{pH} = - \lg [H^+], \text{ thus } [H^+] = 10^{-4.14} = 7.24 \times 10^{-5};$$ $$K_a = \frac{7.24 \times 10^{-5} \times 0.29}{0.281} = 7.48 \times 10^{-5}$$

Answer

$7.48 \times 10^{-5}$ is the ionization constant of HA.

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