An ammonia-ammonium chloride buffer has a pH value of 9 with . What will be the exact new pH if of i — Ionic Equilibrium Chemistry Question
Question
An ammonia-ammonium chloride buffer has a pH value of 9 with $[NH_3] = 0.25 \text{ M}$. What will be the exact new pH if $500 \text{ mL }$ of $0.1 \text{ M } KOH$ is added to $200 \text{ mL }$ of this buffer solution? ($K_b$ for $NH_3 = 2 \times 10^{-5}$ and $\log 2 = 0.3$)
💡 Solution & Explanation
Initial buffer $pH = 9$, so $pOH = 5$. $pK_b = -\log(2 \times 10^{-5}) = 4.7$. By Henderson's equation: $5 = 4.7 + \log([NH_4^+]/0.25)$, giving $\log([NH_4^+]/0.25) = 0.3 = \log 2$. Thus, $[NH_4^+] = 0.5 \text{ M}$. In $200 \text{ mL}$, initial millimoles: $NH_3 = 200 \times 0.25 = 50 \text{ mmol}$, $NH_4^+ = 200 \times 0.5 = 100 \text{ mmol}$. We add $500 \text{ mL}$ of $0.1 \text{ M } KOH$, providing $50 \text{ mmol}$ of $OH^-$. Reaction: $NH_4^+ + OH^- \rightarrow NH_3 + H_2O$. New millimoles: $NH_4^+ = 100 - 50 = 50 \text{ mmol}$, $NH_3 = 50 + 50 = 100 \text{ mmol}$. New $pOH = pK_b + \log(50/100) = 4.7 + \log(0.5) = 4.7 - 0.3 = 4.4$. New $pH = 14 - 4.4 = 9.6$.