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

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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