Calculate the percentage hydrolysis of NaOCN | vedclass

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

Question

(A) The salt $NaOCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HOCN)$.For such a salt,the degree of hydrolysis $h$ is given by the formula

Vedclass · Accepted Answer

$0.01$

Vedclass · Answer

(A) The salt $NaOCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HOCN)$.For such a salt,the degree of hydrolysis $h$ is given by the formula: $h = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{K_w}{K_a 	imes C}}$.Given: $K_w = 10^{-14}$,$K_a = 3.33 	imes 10^{-4}$,and $C = 0.003 \ M = 3 	imes 10^{-3} \ M$.Substituting the values: $h = \sqrt{\frac{10^{-14}}{3.33 	imes 10^{-4} 	imes 3 	imes 10^{-3}}} = \sqrt{\frac{10^{-14}}{10^{-6}}} = \sqrt{10^{-8}} = 10^{-4}$.The percentage hydrolysis is $h 	imes 100 = 10^{-4} 	imes 100 = 0.01 \%$.

Calculate the percentage hydrolysis of $NaOCN$ in its $0.003 \ M$ aqueous solution ($K_a$ for $HOCN = 3.33 \times 10^{-4} \ M$) .....$\%$

Similar Questions

Calculate the hydrolysis constant $(K_h)$ of the salt containing $NO_2^-$. Given the $K_a$ for $HNO_2 = 4.5 \times 10^{-10}$.

If $pK_b$ for $CN^{-}$ at $25\,^{\circ}C$ is $3.7$,what is the $pH$ of a $0.5\, M$ aqueous $NaCN$ solution?

If the value of $K_b$ is $10^{-5}$,find the $K_h$ value for a salt of (strong acid - weak base) at $363 \ K \ (90 \ ^\circ C)$. [At $90 \ ^\circ C, K_w = 10^{-12}$]

The $pH$ of $0.1 \, M$ aqueous solutions of $NaCl$,$CH_3COONa$ and $NH_4Cl$ will follow the order:

The $pK_a$ of a weak acid,benzoic acid and $pK_b$ of a weak base,ammonium hydroxide are $4.25$ and $4.75$ respectively. Then the $pH$ of $0.1 \ M$ solution of ammonium benzoate will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module