Embibe Experts Solutions for Chapter: Equilibrium, Exercise 3: EXERCISE-3

Calculate the $\mathrm{pH}$ of a solution prepared by mixing $50.0\mathrm{mL}$ of $0.200\mathrm{M} {\mathrm{CH}}_3\mathrm{COOH}$ and $50.0\mathrm{mL}$ of $0.100\mathrm{M} \mathrm{NaOH}$. $\left[{\mathrm{K}}_{\mathrm{a}\left({\mathrm{CH}}_3\mathrm{COOH}\right)}=1.8\times {10}^{-5}\right]$

$50 \mathrm{mL}$ of $0.1\mathrm{M} \mathrm{NaOH}$ is added to $75\mathrm{mL}$ of $0.1\mathrm{M} {\mathrm{NH}}_4\mathrm{Cl}$ to make a basic buffer. If ${\mathrm{pK}}_{\mathrm{a}}$ of ${\mathrm{NH}}_4^{+}$ is $9.26$, salculate $\mathrm{pH}$.

A certain solution has a hydrogen ion concentration $4\times {10}^{-3}\mathrm{M}$. For the indicator thymol blue. $\mathrm{pH}$ is $2.0$ when half the indicator is in unionised from. Find the $\%$ of indicator in unionised form in the solution with $\left[{\mathrm{H}}^{+}\right]=4\times {10}^{-3}\mathrm{M}$.

What is the $\mathrm{pH}$ of $0.1\mathrm{M} {\mathrm{NaHCO}}_3$? ${\mathrm{K}}_1=4.5\times {10}^{-7}, {\mathrm{K}}_2=4.5\times {10}^{-11}$ for carbonic acids.

Calculate the hydronium ion concentration and $\mathrm{pH}$ at the equivalence point in the reaction of $22.0 \mathrm{mL}$ of $0.10\mathrm{M}$ acetic acid, ${\mathrm{CH}}_3\mathrm{COOH}$, with $22.0 \mathrm{mL}$ of $0.10\mathrm{M} \mathrm{NaOH}$.

Calculate the hydronium ion concentration and the $\mathrm{pH}$ at the equivalence point in a titration of $50.0 \mathrm{mL}$ of $0.40\mathrm{M} {\mathrm{NH}}_3$ with $0 0 \mathrm{M} \mathrm{HCl}$.

$50\mathrm{ml}$ of a solution which is $0.050\mathrm{M}$ in the acid $\mathrm{HA}\left({\mathrm{pK}}_{\mathrm{a}}=3.80\right)$ and $0.10\mathrm{M}$ in $\mathrm{HB}\left({\mathrm{pK}}_{\mathrm{a}}=8.20\right)$ is titrated with $0.2\mathrm{M} \mathrm{NaOH}$. Calculate the $\mathrm{pH}$ at the first equivalence point, but fill the answer in $\mathrm{OMR}$ sheet by multiplying it with '$1000$'. (i.e. say $\mathrm{pH}$ is $2.\mathrm{8\; then\; mark\; it\; as }$)

The emf of the cell $\mathrm{Ag}\left|\mathrm{AgI}\right|\mathrm{KI}(0.05 \mathrm{M})\left|\right|{\mathrm{AgNO}}_3(0.05 \mathrm{M})\mid \mathrm{Ag}$ is $0.788 \mathrm{V}$. Calculate the solubility product of $\mathrm{AgII}$. If your answer is $\mathrm{x}\times {10}^{-16}$ then $\mathrm{x}$ is: