Observe the following solutions (i) $1 \mathrm{~L}$ of $10^{-6} \mathrm{M} \mathrm{Ag} \mathrm{NO}_3$ (ii)…
Observe the following solutions
(i) $1 \mathrm{~L}$ of $10^{-6} \mathrm{M} \mathrm{Ag} \mathrm{NO}_3$
(ii) $1 \mathrm{~L}$ of $10^{-7} \mathrm{M} \mathrm{AgNO}_3$
(iii) $1 \mathrm{~L}$ of $10^{-9} \mathrm{M} \mathrm{Ag} \mathrm{NO}_3$
(iv) $1 \mathrm{~L}$ of $10^{-3} \mathrm{M} \mathrm{AgNO}_3$
(v) $1 \mathrm{~L}$ of $10^{-5} \mathrm{M} \mathrm{NaCl}$
Which of the above two solutions when mixed will give a white precipitate, $\mathrm{AgCl}$ ?
$\left(\right.$ Given $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{AgCl}=1 \times 10^{-10}$ )
(i), (v)
(ii), (v)
(iv), (v)
(iii), (v)
Solution
A solution that has the concentration of $\mathrm{Ag}^{+}$or $\mathrm{Cl}^{-}$ greater than or equal to the solubility of $\mathrm{Ag}^{+}$or $\mathrm{Cl}^{-}$from the value of $K_{s p}$, will precipitate out.
$\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=(\mathrm{s})(\mathrm{s})=\mathrm{s}^2=1 \times 10^{-10} \mathrm{M}^2 \\ & \Rightarrow \mathrm{s}=10^{-5} \mathrm{M}\end{aligned}$
Thus, (iv) $1 \mathrm{~L}$ of $10^{-3} \mathrm{M} \mathrm{AgNO}_3$ and (v) $1 \mathrm{~L}$ of $10^{-5} \mathrm{M}$ $\mathrm{NaCl}$ will precipitate