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Exercises · 6.68

Q.The solubilit y product constant of Ag 2CrO4 and AgBr are 1.1 × 10⁻¹² and 5.0 × 10⁻¹³ respectively. Calculate the ratio of the molarities of their saturated solutions.

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Solving each solubility product for the molar solubility ss gives s(Ag2CrO4)=6.5×10−5 Ms(\text{Ag}_2\text{CrO}_4) = 6.5 \times 10^{-5}\ \text{M} and s(AgBr)=7.07×10−7 Ms(\text{AgBr}) = 7.07 \times 10^{-7}\ \text{M}, so the ratio of molarities is about 92:192 : 1.

Solution

Ag2CrO4\text{Ag}_2\text{CrO}_4 dissolves as Ag2CrO4→2Ag++CrO42−\text{Ag}_2\text{CrO}_4 \rightarrow 2\text{Ag}^+ + \text{CrO}_4^{2-}, so if solubility is s1s_1:

Ksp=(2s1)2(s1)=4s13K_{sp} = (2s_1)^2(s_1) = 4s_1^3

s1=(Ksp4)1/3=(1.1×10−124)1/3=(2.75×10−13)1/3=6.5×10−5 Ms_1 = \left(\frac{K_{sp}}{4}\right)^{1/3} = \left(\frac{1.1 \times 10^{-12}}{4}\right)^{1/3} = \left(2.75 \times 10^{-13}\right)^{1/3} = 6.5 \times 10^{-5}\ \text{M}

AgBr\text{AgBr} dissolves as AgBr→Ag++Br−\text{AgBr} \rightarrow \text{Ag}^+ + \text{Br}^-, so if solubility is s2s_2: …

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