If the pK_b for CN^- is 4.7 at 25,^oC,then th | vedclass

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

Question

(C) $NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$.For the salt of a weak acid and strong base,the $pH$ is given by the formula:$p

Vedclass · Accepted Answer

$11.5$

Vedclass · Answer

(C) $NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$.For the salt of a weak acid and strong base,the $pH$ is given by the formula:$pH = 7 + \frac{1}{2} pK_a + \frac{1}{2} \log C$Given $pK_b(CN^-) = 4.7$,we know $pK_a(HCN) = 14 - 4.7 = 9.3$.Concentration $C = 0.5\,M = 5 	imes 10^{-1}\,M$.Substituting the values:$pH = 7 + \frac{1}{2}(9.3) + \frac{1}{2} \log(5 	imes 10^{-1})$$pH = 7 + 4.65 + \frac{1}{2}(\log 5 + \log 10^{-1})$$pH = 11.65 + \frac{1}{2}(0.699 - 1)$$pH = 11.65 + \frac{1}{2}(-0.301)$$pH = 11.65 - 0.15 = 11.5$.

If the $pK_b$ for $CN^-$ is $4.7$ at $25\,^oC$,then the $pH$ of a $0.5\,M \,NaCN$ solution is:

Similar Questions

Find the equilibrium constant for the equilibrium reaction:
$HCOO^{-} + H_2O \rightleftharpoons HCOOH + OH^{-}$
Given that the dissociation constant of formic acid is $K_a(HCOOH) = 1.8 \times 10^{-4}$.

The $pH$ of a $0.01 \, M$ ammonium cyanide $(NH_4CN)$ solution is $7.02$. If $5 \, L$ of water is added to this solution,the $pH$ will be .......

An aqueous solution of $CH_3COONa$ will be

The aqueous solution of which of the following will have the highest $pH$?

Which of the following salts will undergo hydrolysis when dissolved in water?

Vedclass Test Series

Exam Paper Generator

Online Exam Module