Ncert Exercises & Solutions

7.68 The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and
5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Let s be the solubility of {Ag}_{2} {CrO}_{4} .

Then, {Ag}_{g} {CrO}_{4} \leftrightarrow {Ag}^{2 +} + 2 {CrO}_{4}^{-}

K_{sp} = {(2 s)}^{2} . s = 4 s^{3}

1.1 \times 10^{- 10} = 4^{s}

s = 6.5 \times 10^{- 5} M

Let s' be the solubility of AgBr .

{AgBr}_{(s)} \leftrightarrow {Ag}^{+} + {Br}^{-}

K_{sp} = S'^{2} = 5.0 \times 10^{- 13}

∴ s' = 7.07 \times 10^{- 7} M

Therefore, the ratio of the molarities of their saturated solution is

\frac{2}{s'} = \frac{6.5 \times 10^{- 5} M}{7.07 \times 10^{- 7} M} = 91.9 .

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