Question 64 of 69intermediate🔧 ApplyNumerical3 marks

The solubility product constant of Ag2_2CrO4_4 and AgBr are 1.1×10−121.1 \times 10^{-12} and 5.0×10−135.0 \times 10^{-13} respectively. Calculate the ratio of the molarities of their saturated solutions.

Correct Answer

The ratio of the molarities of the saturated solutions of Ag2_2CrO4_4 and AgBr is approximately 92.

Exercise: EXERCISES | Q: 6.68 | (Page 46)
For More Understanding

Explanation

To find the ratio of the molarities of their saturated solutions, we first need to calculate the molar solubility (SS) of each salt using their respective solubility product constants (KspK_{sp}). For a salt MxXyM_xX_y, the solubility product is given by Ksp=(xS)x(yS)yK_{sp} = (xS)^x(yS)^y. Ag2_2CrO4_4 dissociates into 2 ions of Ag+^+ and 1 ion of CrO42−_4^{2-}, making its Ksp=4S3K_{sp} = 4S^3. AgBr dissociates into 1 ion of Ag+^+ and 1 ion of Br−^-, making its Ksp=S2K_{sp} = S^2. After finding the individual solubilities, their ratio is calculated.

Solution Steps

  1. Step 1: Calculate the molar solubility (SS) of Ag2_2CrO4_4 For Ag2CrO4⇌2Ag++CrO42−Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}, the solubility product is Ksp=(2S)2(S)=4S3K_{sp} = (2S)^2(S) = 4S^3. Given Ksp=1.1×10−12K_{sp} = 1.1 \times 10^{-12}, we have:

    4S3=1.1×10−124S^3 = 1.1 \times 10^{-12} S3=1.1×10−124=0.275×10−12S^3 = \frac{1.1 \times 10^{-12}}{4} = 0.275 \times 10^{-12} S=(0.275×10−12)1/3=0.651×10−4S = (0.275 \times 10^{-12})^{1/3} = 0.651 \times 10^{-4}

    S=6.51×10−5S = 6.51 \times 10^{-5} M

  2. Step 2: Calculate the molar solubility (SS) of AgBr For AgBr⇌Ag++Br−AgBr \rightleftharpoons Ag^+ + Br^-, the solubility product is Ksp=(S)(S)=S2K_{sp} = (S)(S) = S^2. Given Ksp=5.0×10−13K_{sp} = 5.0 \times 10^{-13}, we have:

    S2=5.0×10−13S^2 = 5.0 \times 10^{-13} S=5.0×10−13=50×10−14S = \sqrt{5.0 \times 10^{-13}} = \sqrt{50 \times 10^{-14}}

    S=7.07×10−7S = 7.07 \times 10^{-7} M

  3. Step 3: Calculate the ratio of their molarities Ratio = Molarity of Ag2CrO4Molarity of AgBr\frac{\text{Molarity of } Ag_2CrO_4}{\text{Molarity of } AgBr} Ratio = 6.51×10−57.07×10−7=0.92×102=92\frac{6.51 \times 10^{-5}}{7.07 \times 10^{-7}} = 0.92 \times 10^2 = 92

PreviousNext