0.5moldm³ 30cm³ CH₃COOH solution is neutralized by Ba(OH)₂ solution. 0.75dm³ of the solution is used. If 10cm³ of Ba(OH)₂ solution is neutralized by 0.1moldm⁻³ NaOH solution, calculate W.

Initial amount of $\mathrm{CH_3COOH}$:

$$n = 0.50 \, \mathrm{mol \, dm^{-3}} \times 0.025 \, \mathrm{dm^3} = 0.0125 \, \mathrm{mol}$$

Equivalence volume of $\mathrm{NaOH}$:

$$V_{\mathrm{eq}} = \frac{0.0125 \, \mathrm{mol}}{0.10 \, \mathrm{mol \, dm^{-3}}} = 0.125 \, \mathrm{dm^3} = 125 \, \mathrm{cm^3}$$

$K_a = 1.8 \times 10^{-5} \, \mathrm{mol \, dm^{-3}}$
$\mathrm{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74$

(i) Before adding $\mathrm{NaOH}$
Weak acid:

$$[\mathrm{H^+}] = \sqrt{K_a C}$$

Substitute:

$$[\mathrm{H^+}] = \sqrt{1.8 \times 10^{-5} \times 0.50} = \sqrt{9.0 \times 10^{-6}} = 3.0 \times 10^{-3} \, \mathrm{mol \, dm^{-3}}$$ $$\mathrm{pH} = -\log(3.0 \times 10^{-3}) = 2.52$$

2.52

(ii) After adding $10 \, \mathrm{cm^3}$ $\mathrm{NaOH}$
Moles $\mathrm{NaOH}$ added:

$$n = 0.10 \, \mathrm{mol \, dm^{-3}} \times 0.010 \, \mathrm{dm^3} = 0.0010 \, \mathrm{mol}$$

Moles $\mathrm{CH_3COOH}$ left:

$$0.0125 - 0.0010 = 0.0115 \, \mathrm{mol}$$

Moles $\mathrm{CH_3COO^-}$:

$$0.0010 \, \mathrm{mol}$$

Total volume:

$$25 + 10 = 35 \, \mathrm{cm^3} = 0.035 \, \mathrm{dm^3}$$

Buffer:

$$\mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{n_{\mathrm{CH_3COO^-}}}{n_{\mathrm{CH_3COOH}}} \right)$$

Ratio:

$$\frac{0.0010}{0.0115} = 0.087$$ $$\log(0.087) = -1.06$$ $$\mathrm{pH} = 4.74 + (-1.06) = 3.68$$

3.68

(iii) At equivalence point ($125 \, \mathrm{cm^3}$ $\mathrm{NaOH}$)
Moles $\mathrm{CH_3COO^-} = 0.0125 \, \mathrm{mol}$
Total volume:

$$25 + 125 = 150 \, \mathrm{cm^3} = 0.150 \, \mathrm{dm^3}$$ $$C_b = \frac{0.0125}{0.150} = 0.0833 \, \mathrm{mol \, dm^{-3}}$$ $$K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \, \mathrm{mol \, dm^{-3}}$$

Weak base:

$$[\mathrm{OH^-}] = \sqrt{K_b C_b}$$

Substitute:

$$[\mathrm{OH^-}] = \sqrt{5.56 \times 10^{-10} \times 0.0833} = \sqrt{4.63 \times 10^{-11}} = 6.8 \times 10^{-6} \, \mathrm{mol \, dm^{-3}}$$ $$\mathrm{pOH} = -\log(6.8 \times 10^{-6}) = 5.17$$ $$\mathrm{pH} = 14.00 - 5.17 = 8.83$$

8.83

(iv) After adding $30 \, \mathrm{cm^3}$ $\mathrm{NaOH}$
Moles $\mathrm{NaOH}$ added:

$$0.10 \times 0.030 = 0.0030 \, \mathrm{mol}$$

Moles $\mathrm{CH_3COOH}$ left:

$$0.0125 - 0.0030 = 0.0095 \, \mathrm{mol}$$

Moles $\mathrm{CH_3COO^-}$:

$$0.0030 \, \mathrm{mol}$$

Total volume:

$$25 + 30 = 55 \, \mathrm{cm^3} = 0.055 \, \mathrm{dm^3}$$

Buffer:

$$\frac{n_{\mathrm{CH_3COO^-}}}{n_{\mathrm{CH_3COOH}}} = \frac{0.0030}{0.0095} = 0.316$$ $$\log(0.316) = -0.50$$ $$\mathrm{pH} = 4.74 + (-0.50) = 4.24$$

4.24