Ramendra C Mukerjee Solutions for Chapter: Ionic Equilibrium in Aqueous Solutions, Exercise 1: PROBLEMS

Calculate the simultaneous solubility of ${\mathrm{CaF}}_2$ and ${\mathrm{SrF}}_2$.

${\mathrm{K}}_{\mathrm{sp}} \left({\mathrm{CaF}}_2\right)=3.9\times {10}^{-11}$

${\mathrm{K}}_{\mathrm{sp} }\left({\mathrm{SrF}}_2\right)=2.9\times {10}^{-9}$

Aniline is a weak organic base in aqueous solutions. Suggest a solvent in which aniline would become a strong base.

Distinguish between acid strength and acid concentration.

Liquid ${\mathrm{NH}}_3$ like water, is an amphiprotic solvent. Write the equation for the auto-ionisation of ${\mathrm{NH}}_3$.

Calculate the sulphate ion concentration in $0.15 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$. Given: ${\mathrm{K}}_2=1.02\times {10}^{-2}$.

If the answer is of type $\mathrm{y}\times {10}^{-3} \mathrm{M}$, report the value of $\mathrm{y}$ correct up to two places of decimals.

A $50.0 \mathrm{mL}$ sample of a $0.01 \mathrm{M}$ solution of $\mathrm{HCOOH}$ was titrated with $0.10 \mathrm{M} \mathrm{NaOH}$. Calculate the $\mathrm{pH}$ of the solution when $10 \mathrm{mL}$ of $\mathrm{NaOH}$ was added. Given: ${\mathrm{K}}_{\mathrm{a}}$ for $\mathrm{HCOOH}=1.772\times {10}^{-4}$

What is the $\mathrm{pH}$ of a $0.10 \mathrm{M}$ solution of ethylenediaminetetraacetic acid $\left(\mathrm{EDTA}\right)$?

${\mathrm{K}}_{{\mathrm{a}}_1}=1\times {10}^{-2}, {\mathrm{K}}_{{\mathrm{a}}_2}=2.1\times {10}^{-3}, {\mathrm{K}}_{{\mathrm{a}}_3}=6.9\times {10}^{-7}$ and ${\mathrm{K}}_{{\mathrm{a}}_4}=5.5\times {10}^{-11}$.

A concentrated strong acid is added to a solid mixture of $0.015 \mathrm{mole}$ sample of $\mathrm{Fe}(\mathrm{OH}{)}_2$ and $\mathrm{Cu}(\mathrm{OH}{)}_2$, placed in one litre of water. At what value of $\mathrm{pH}$ will the dissolution of each hydroxide be complete? (Assume negligible volume change)

${\mathrm{K}}_{\mathrm{sp} }\left[\mathrm{Fe}(\mathrm{OH}{)}_2\right]=7.9\times {10}^{-15}$ and ${\mathrm{K}}_{\mathrm{sp} }\left[\mathrm{Cu}(\mathrm{OH}{)}_2\right]=1.6\times {10}^{-19}$

Answer correct up to two places of decimal.