The solubility product of ${\mathrm{BaF}}_2$ has the value $1.7\times {10}^{-4}$. If ${\mathrm{V}}_1\mathrm{mL}$ of ${\mathrm{BaCI}}_2$ solution $\left({10}^{-1}\mathrm{M}\right)$ and ${\mathrm{V}}_2\mathrm{mL} \mathrm{of} \mathrm{NaF}$ solution $\left({10}^{-2}\mathrm{M}\right)$ are mixed,what is the ratio ${\mathrm{V}}_1:{\mathrm{V}}_2$ so that a precipitate may be obtained?

Statement-1: To get a buffer solution at maximum capacity the molar ratio of ${\mathrm{CH}}_3\mathrm{COOH} \mathrm{and} \mathrm{NaOH}$ should be $2:1$

Statement-2: The ionic conductivity of ${\mathrm{A}}^{2+} \mathrm{and} {\mathrm{B}}^{-}$ at infinite dilution are $100 \mathrm{and} 50{\mathrm{Scm}}^2{\mathrm{eq}}^{-1}$ respectively. The equivalent conductivity of ${\mathrm{AB}}_2$ at infinite dilution would be $200\mathrm{S} {\mathrm{cm}}^2{\mathrm{eq}}^{-1}$

What volume of $0.1\mathrm{M} {\mathrm{RNH}}_3\mathrm{CI}$ solution should br added to $100\mathrm{ml}$ of $0.1\mathrm{M} {\mathrm{RNH}}_2$ solution to produce a buffer solution of $\mathrm{pH}=8.7?$

Given: ${\mathrm{pK}}_{\mathrm{b}}\mathrm{of} {\mathrm{RNH}}_2=5; \log 2=0.3$

$1$ mole of ${\mathrm{AgNO}}_3$ added to $100\mathrm{litre} \mathrm{of} 1\mathrm{M} {\mathrm{NH}}_3$. What is the concentration of $\mathrm{Ag}{{\left({\mathrm{NH}}_3\right)}_2}^{+}$ in solution?

[Given \; for $\mathrm{Ag}{{\left({\mathrm{NH}}_3\right)}_2}^{+}, {\mathrm{K}}_{{\mathrm{f}}_1}=2.0\times {10}^3;{\mathrm{K}}_{{\mathrm{f}}_2}={10}^4$]