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.

Asked by Pragya Singh | 1 year ago |  184

1 Answer

Solution :-

Ag2CrO4→ 2Ag2+ + \( CrO_{4}^{-}\)

Now, Ksp = [Ag2+]2\( [CrO_{4}^{-}]\)

Assuming the solubility of Ag2CrO4 is ‘x’.

Thus,[Ag2+] = 2x and \( CrO_{4}^{-}\)​ = x

Ksp = \( (2x)^2×x\)

\( 1.1\times 10^{-12} = 4x^{3}\)

\( x = 0.65\times 10^{-4}M\)

Assuming the solubility of AgBr is y.

AgBr(s) → Ag2+ + Br

Ksp = (y)2

\( 5.0\times 10^{-13} = y^{2}\)

y = \( 7.07\times 10^{-7}M\)

The ratio of molarities to their saturated solution is:

\( \dfrac{x}{y} = \dfrac{0.65\times10^{-4}M}{7.07\times10^{-7}M} = 91.9\)

Answered by Abhisek | 1 year ago

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