The dissociation constant of HCN is 5 times 1 | vedclass

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

Question

(C) The given solution is a buffer solution consisting of a weak acid $(HCN)$ and its salt with a strong base $(KCN)$.Using the Henderson-Hasselbalch

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) The given solution is a buffer solution consisting of a weak acid $(HCN)$ and its salt with a strong base $(KCN)$.Using the Henderson-Hasselbalch equation:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given:$K_a = 5 	imes 10^{-10}$$pK_a = - \log(5 	imes 10^{-10}) = 10 - \log 5 = 10 - 0.698 = 9.302$Concentration of salt $(KCN)$ = $\frac{0.15 \, mole}{0.5 \, dm^3} = 0.3 \, M$Concentration of acid $(HCN)$ = $\frac{1.5 \, mole}{0.5 \, dm^3} = 3.0 \, M$$pH = 9.302 + \log \left( \frac{0.3}{3.0} ight)$$pH = 9.302 + \log(0.1)$$pH = 9.302 - 1 = 8.302$

The dissociation constant of $HCN$ is $5 \times 10^{-10}$. The $pH$ of the solution prepared by mixing $1.5 \, mole$ of $HCN$ and $0.15 \, mole$ of $KCN$ in water and making up the total volume to $0.5 \, dm^3$ is (in $.302$)

Similar Questions

Equal volumes of $0.5 \ N$ acetic acid and $0.5 \ N$ sodium acetate are mixed. What is the $pH$ of the resultant solution? $(pK_{a}$ of acetic acid $= 4.75)$

At $298 \text{ K}$, a certain buffer solution contains equal concentrations of $X^-$ and $HX$. The $K_b$ for $X^-$ is $10^{-10}$. What is the $pH$ of this buffer solution?

$20 \ mL$ of $0.1 \ M$ acetic acid is mixed with $50 \ mL$ of potassium acetate. $K_a$ of acetic acid $= 1.8 \times 10^{-5}$ at $27^{\circ} C$. Calculate the concentration of potassium acetate if the $pH$ of the mixture is $4.8$. (in $M$)

Which of the following will form a basic buffer solution?

If the molar concentrations of a base and its conjugate acid are the same, then the $pOH$ of the buffer solution is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module