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Increasing amount of dilute HCl was added to an aqueous solution containing $10^{- 4} M$ each of ${Pb}^{+ 2}$ and ${Hg}_{2}^{+ 2}$ . Given that $K_{sp}$ value of ${PbCl}_{2}$ and ${Hg}_{2} {Cl}_{2}$ are $10^{- 8}$ and $10^{- 16}$ respectively at $25 ° C$ then value of pH of the solution at which precipitation of the least soluble metal halide will start is:

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answer is 6.

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Detailed Solution

Least soluble metal chloride out of the given two is ${Hg}_{2} {Cl}_{2} (Ksp = 10^{- 16})$
To just start precipitation of ${Hg}_{2} {Cl}_{2}$

let the concentration of HCl required is “C”, such that $10^{- 16} = 10^{- 4} {[C]}^{2}$
(or) $C^{2} = \frac{10^{- 16}}{10^{- 4}} = 10^{- 12}$ (or) $C = 10^{- 6} M = [H^{+}]$

$∴$ pH of the solution = 6

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