A buffer solution is prepared in which the concentration of ${\rm{N}}{{\rm{H}}_3}$ is $0.30\;M$ and the concentration of ${\rm{NH}}_4^ + $ is $0.20\;M$. If the equilibrium constant, ${K_b}$ for ${\rm{N}}{{\rm{H}}_3}$ equals $1.8 \times {10^{ - 5}}$ what is the pH of this solution? $(\log 2.7 = 0.43)$

JEE CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

A buffer solution is prepared in which the concentration of

\mathrm{N}{\mathrm{H}_3}

0.30\;M

and the concentration of

\mathrm{NH}_4^ +

0.20\;M

. If the equilibrium constant,

{K_b}

for

\mathrm{N}{\mathrm{H}_3}

equals

1.8 \times {10^{ - 5}}

what is the pH of this solution?

(\log 2.7 = 0.43)

View answer and explanation

Correct option: C

9.43

Option C (

9.43

) is correct because it is calculated using the Henderson-Hasselbalch equation for a basic buffer solution.

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\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