Chemistry - Ionic Equilibrium Question with Solution | TestHub

In Mohr's method, solution of ${\mathrm{Cl}}^{-}$is titrated against standard ${\mathrm{AgNO}}_3$. During the titration, a small quantity of ${\mathrm{K}}_2{\mathrm{CrO}}_4$ solution is added as an indicator. As AgCl is less soluble than ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$, the added ${\mathrm{Ag}}^{+}$reacts first with ${\mathrm{Cl}}^{-}$than with ${\mathrm{CrO}}_4{}^{2-}$.

At the equivalence point, when ${\mathrm{Cl}}^{-}$is precipitated quantitatively as AgCl , the slight excess of ${\mathrm{Ag}}^{+}$reacts with ${\mathrm{CrO}}_4{}^{2-}$ to give intensely red precipitate of ${\mathrm{Ag}}_2{\mathrm{CrO}}_4\cdot 50.0\mathrm{mL}$ of $0.1\mathrm{M}{\mathrm{Cl}}^{-}$solution is titrated against 0.1 M standard ${\mathrm{AgNO}}_3$.

What must be the theoretical molar concentration of ${\mathrm{CrO}}_4{}^{2-}$ in the solution so that ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$ will start precipitating at the equivalence-point. Given : ${\mathrm{K}}_{\mathrm{sp}}(\mathrm{AgCl})=10^{-10}{\mathrm{M}}^2;{\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)=4\times 10^{-12}{\mathrm{M}}^3$

[Fill your answer by multiplying it with 100.]