At 25 , ^circ C,the K_a for HCN is 5 times 10 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The $pH$ of a buffer solution is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.First,calculate $pK_a$: $pK

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The $pH$ of a buffer solution is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(5 	imes 10^{-10}) = 10 - \log(5) = 10 - 0.699 = 9.301 \approx 9.3$.Given $pH = 9$,$[Acid] = [HCN] = 2 \, M$,$V_{acid} = 10 \, mL$,$[Salt] = [KCN] = 5 \, M$,and $V_{salt} = V$.The concentration of salt in the mixture is $\frac{5V}{10+V}$ and acid is $\frac{2 	imes 10}{10+V}$.Substituting into the equation: $9 = 9.3 + \log \left( \frac{5V}{20} ight)$.$-0.3 = \log \left( \frac{V}{4} ight)$.$10^{-0.3} = \frac{V}{4}$.$0.5 = \frac{V}{4} \implies V = 2 \, mL$.

At $25 \, ^\circ C$,the $K_a$ for $HCN$ is $5 \times 10^{-10}$. What volume (in $mL$) of $5 \, M \, KCN$ solution must be added to $10 \, mL$ of $2 \, M \, HCN$ solution to maintain a constant $pH$ of $9$?

Similar Questions

$A$ buffer solution can be prepared by mixing equal volumes of

Buffer system helps to maintain blood $pH$ between .......

$100 \ mL$ of a solution contains $0.1 \ M$ $NH_4OH$ and $0.1 \ M$ $NH_4Cl$. The $pH$ of the solution will not change on adding :-

Calculate the $pH$ of a $0.10 \,M$ ammonia solution. Calculate the $pH$ after $50.0 \,mL$ of this solution is treated with $25.0 \,mL$ of $0.10 \,M$ $HCl$. The dissociation constant of ammonia,$K_{b} = 1.77 \times 10^{-5}$.

What are buffer solutions? Explain types of buffer solutions with examples.

Vedclass Test Series

Exam Paper Generator

Online Exam Module