The ionization constant of HCN is $4.5{\rm{ }} \times {\rm{ }}{10^{ - 10}}$ at 298 K. Hydrolysis constant of sodium cyanide is

$2.22{\rm{ }} \times {\rm{ }}{10^{--5}}$

$1.11{\rm{ }} \times {\rm{ }}{10^{--6}}$

$3.33{\rm{ }} \times {\rm{ }}{10^{ - 4}}$

$4.44{\rm{ }} \times {\rm{ }}{10^{ - 7}}$

JEE CHEMISTRY MCQ: Equilibrium

1 Medium

1 of 1

Application +4 | -1

The ionization constant of HCN is

4.5\mathrm{ } \times \mathrm{ }{10^{ - 10}}

at 298 K. Hydrolysis constant of sodium cyanide is

View answer and explanation

Correct option: C

2.22\mathrm{ } \times \mathrm{ }{10^{--5}}

Option C (2.22 x 10⁻⁵) is correct because it is the hydrolysis constant of sodium cyanide calculated using the relationship between Kw, Ka, and Kh.

Practice All Chapter MCQs →

\mathrm{K}_1

\mathrm{K}_2

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

K{a_1},\,\,K{a_2}\,,K{a_3}

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

\text{O}_2