Exactly of is systematically added to of . The neutralization mechanism forces the strong base to co — Ionic Equilibrium Chemistry Question
Question
Exactly $50 \text{ mL}$ of $0.1 \text{ M } NaOH$ is systematically added to $75 \text{ mL}$ of $0.1 \text{ M } NH_4Cl$. The neutralization mechanism forces the strong base to consume the weak acid salt, producing a basic buffer. Given that $pK_a$ for the conjugate acid $NH_4^+$ is $9.26$, calculate the precise final pH of the resulting mixture. (Use $\log 2 = 0.30$).
💡 Solution & Explanation
Initial millimoles of $NaOH$ = $5 \text{ mmol}$. Initial millimoles of $NH_4Cl$ = $7.5 \text{ mmol}$. The mechanism is $NH_4^+ + OH^- \rightarrow NH_3 + H_2O$. $5 \text{ mmol}$ of $OH^-$ will completely consume $5 \text{ mmol}$ of $NH_4^+$, generating exactly $5 \text{ mmol}$ of $NH_3$. The unreacted $NH_4^+$ remaining is $7.5 - 5 = 2.5 \text{ mmol}$. The buffer consists of $5 \text{ mmol}$ Base ($NH_3$) and $2.5 \text{ mmol}$ Salt ($NH_4^+$). Since $pK_a(NH_4^+) = 9.26$, $pK_b(NH_3) = 14 - 9.26 = 4.74$. Applying Henderson-Hasselbalch: $pOH = pK_b + \log\frac{[Salt]}{[Base]} = 4.74 + \log(2.5 / 5) = 4.74 + \log(0.5) = 4.74 - 0.30 = 4.44$. The final pH = $14 - 4.44 = 9.56$.