A basic buffer is perfectly formed by systematically mixing of with of . Assuming the exact of the c — Ionic Equilibrium Chemistry Question
Question
A basic buffer is perfectly formed by systematically mixing $50 \text{ mL}$ of $0.1 \text{ M } NaOH$ with $75 \text{ mL}$ of $0.1 \text{ M } NH_4Cl$. Assuming the exact $pK_a$ of the conjugate acid $NH_4^+$ is exactly $9.26$ at the working temperature, calculate the precise pH of the resulting buffer solution. (Take $\log 2 = 0.30$).
💡 Solution & Explanation
Total millimoles of $NaOH$ = $50 \times 0.1 = 5 \text{ mmol}$. Total millimoles of $NH_4Cl$ = $75 \times 0.1 = 7.5 \text{ mmol}$. The strong base completely reacts with the weak acid salt: $NH_4Cl + NaOH \rightarrow NH_3 + NaCl + H_2O$. The $5 \text{ mmol}$ of $NaOH$ effectively neutralizes $5 \text{ mmol}$ of $NH_4Cl$ to strictly produce exactly $5 \text{ mmol}$ of the weak base $NH_3$. The unreacted $NH_4Cl$ remaining is $7.5 - 5 = 2.5 \text{ mmol}$. Since $pK_a(NH_4^+) = 9.26$, the $pK_b(NH_3) = 14 - 9.26 = 4.74$. Applying the Henderson-Hasselbalch equation for basic buffers: $pOH = pK_b + \log\frac{[Salt]}{[Base]} = 4.74 + \log\left(\frac{2.5/V}{5.0/V}\right) = 4.74 + \log(0.5) = 4.74 - 0.30 = 4.44$. The final resulting pH is $14.0 - 4.44 = 9.56$.