Three solutions of strong electrolytes, 25 ml of 0.1 M – HX, 25 ml of 0.1 M – and 50 ml of 0.1 N – a — Ionic Equilibrium Chemistry Question
Question
Three solutions of strong electrolytes, 25 ml of 0.1 M – HX, 25 ml of 0.1 M – $\text{H}_2\text{Y}$ and 50 ml of 0.1 N – $\text{Z(OH)}_2$ are mixed. pOH of the resulting solution is
💡 Solution & Explanation
Moles of $\text{H}^+$ from HX = $25 \times 0.1 = 2.5\text{ mmol}$. Moles of $\text{H}^+$ from diprotic $\text{H}_2\text{Y}$ = $25 \times 0.1 \times 2 = 5.0\text{ mmol}$. Total $\text{H}^+ = 7.5\text{ mmol}$. Moles of $\text{OH}^-$ from $\text{Z(OH)}_2$ (given in Normality, which already accounts for the 2 $\text{OH}^-$ ions) = $50 \times 0.1 = 5.0\text{ mmol}$. Excess $\text{H}^+ = 7.5 - 5.0 = 2.5\text{ mmol}$. Total volume = $25 + 25 + 50 = 100\text{ ml}$. Final $[\text{H}^+] = 2.5 / 100 = 0.025\text{ M}$. $\text{pH} = -\log(2.5 \times 10^{-2}) = 2 - \log 2.5 \approx 2 - 0.4 = 1.6$. The $\text{pOH} = 14 - 1.6 = 12.4$.