Ionic EquilibriumhardMCQ SINGLE

for is at . For maintaining a constant pH of , what is the exact volume of solution required to be aIonic Equilibrium Chemistry Question

Question

$K_a$ for $HCN$ is $5 \times 10^{-10}$ at $25^\circ C$. For maintaining a constant pH of $9.0$, what is the exact volume of $5 \text{ M } KCN$ solution required to be added to $10 \text{ mL}$ of $2 \text{ M } HCN$ solution?

Answer: C

💡 Solution & Explanation

$pK_a = -\log(5 \times 10^{-10}) = 10 - \log 5 = 10 - 0.7 = 9.3$. Using the Henderson equation: $pH = pK_a + \log\frac{[Salt]}{[Acid]} \Rightarrow 9.0 = 9.3 + \log\frac{[Salt]}{[Acid]} \Rightarrow \log\frac{[Salt]}{[Acid]} = -0.3$. This implies $\frac{[Salt]}{[Acid]} = 10^{-0.3} = 0.5$. The millimoles of salt = $5 \times V$, and acid = $2 \times 10 = 20$. Therefore, $\frac{5V}{20} = 0.5 \Rightarrow 5V = 10 \Rightarrow V = 2 \text{ mL}$.

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full NEET Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry