Question

Question: The Mohr method is a technique for determining the amount of chloride ion in an unknown sample. It is based on the difference in solubility between silver chloride ${\mathbf{\text{(AgCl;k}}}_{\mathbf{\text{SP}}}{\mathbf{\text{= 1.6×10}}}^{\mathbf{\text{- 10}}}\mathbf{\text{)}}$and silver chromate${\mathbf{\text{(AgCrO}}}_{\mathbf{\text{4}}}{\mathbf{\text{;K}}}_{\mathbf{\text{SP}}}{\mathbf{\text{= 1.9×10}}}^{\mathbf{\text{- 12}}}\mathbf{\text{)}}$ In using this method, one adds a small amount of CI^- chromate ion to a solution with unknown chloride concentration. By measuring the volume of AgNO₃ added before the appearance of the red silver chromate, one can determine the amount of originally present. Suppose we have a solution that is 0.100 M in CI^- and in. If we add 0.100M solution AgNO₃ drop by drop, will AgCI or AgCrO₄ precipitate first? When first appears, what fraction of CI^- the originally present remains in solution?

Short answer

Answer

$5.8\times {10}^{-3}\%$

Step-by-step solution

Find which will precipitate first

We need to calculate concentration ofin each of them, in AgCl:

$$\begin{gathered} \left[A{g}^{+}\right]=\frac{K_{sp}}{\left[C{l}^{-}\right]} \\ \left[A{g}^{+}\right]=\frac{1.6\times {10}^{-10}}{0.1} \\ [A{g}^{}+]=1.6\times {10}^{-9}M \end{gathered}$$

In$A{g}_{2}Cr{O}_4$:

$$\begin{gathered} [A{g}^{+}{]}^2=\frac{K_{SP}}{[CrO\_4^{2-}]} \\ [A{g}^{+}{]}^2=\frac{1.9\times {10}^{-12}}{0.00250} \\ [A{g}^{+}{]}^2=7.6\times {10}^{-10}M \\ [A{g}^{}+]=2.76\times {10}^{-5}M \end{gathered}$$

Since $\left[A{g}^{+}\right]AgCl<\left[A{g}^{+}\right]AgCrO,AgCl$will precipitate first.

Finding fraction

Precipitate when$\left[A{g}^{+}\right]=2.76\times {10}^{-5}M$ . So, the concentration of is then:

$$\begin{gathered} \left[C{l}^{-}\right]=\frac{K_{SP}}{\left[A{g}^{+}\right]} \\ [C{l}^{}-]=\frac{1.6\times {10}^{-10}}{2.76\times {10}^{-5}} \\ [C{l}^{}-]=5.8\times {10}^{-6}M \end{gathered}$$

The fraction of the CI^- originally present that remains in solution is:

$\frac{5.8\times {10}^{-6}}{0.1}\times 100\%=5.8\times {10}^{-3}\%$