The ${\mathrm{K}}_{\mathrm{sp}}$ for bismuth sulphide $\left({\mathrm{Bi}}_2{\mathrm{S}}_3\right)$ is $1.08\times {10}^{-73}$. The solubility of ${\mathrm{Bi}}_2{\mathrm{S}}_3$ in ${\mathrm{molL}}^{-1}$ at $298 \mathrm{K}$ is

The percentage of pyridine (${\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}$) that forms a pyridinium ion (${\mathrm{C}}_5{\mathrm{H}}_5{\mathrm{N}}^{+}\mathrm{H}$) in a $0.10 \mathrm{M}$ aqueous pyridine solution is, (${\mathrm{K}}_{\mathrm{b}} \mathrm{for} {\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}=1.7\times {10}^{-9}$).

Following solutions were prepared by mixing different volumes of $\mathrm{NaOH} \text{and} \mathrm{HCl}$of different concentrations ?

a. $60\mathrm{mL}\frac{\mathrm{M}}{10}\mathrm{HCl}+40\mathrm{mL}\frac{\mathrm{M}}{10}\mathrm{NaOH}$
b. $55\mathrm{mL}\frac{\mathrm{M}}{10}\mathrm{HCl}+45\mathrm{mL}\frac{\mathrm{M}}{10}\mathrm{NaOH}$

c. $75\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{HCl}+25\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{NaOH}$
d. $100\mathrm{mL}\frac{\mathrm{M}}{10}\mathrm{HCl}+100\mathrm{mL}\frac{\mathrm{M}}{10}\mathrm{NaOH}$

The solubility of ${\mathrm{BaSO}}_4$ in water is $2.42\times {10}^{-3} \mathrm{g} {\mathrm{L}}^{-1}$ at $298 \mathrm{K}$. The value of its solubility product $\left({\mathrm{K}}_{\mathrm{sp}}\right)$ will be

(Given molar mass of ${\mathrm{BaSO}}_4$ = $233 \mathrm{g} {\mathrm{L}}^{-1}$ )