Solubility and Equilibrium - AP Chemistry

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Question

Calculate the molar solubility of AgBr in 0.050 M AgNO3 at room temperature. The Ksp of AgBr is 5.4 x 10-13.

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Answer

$K_sp = [Ag^+ ][Br^- ] = 5.4 x 10^{-13}$

$5.4 x 10^{-13} = [0.05+x][x] \sim [0.05][x]$

$x = 1.08 x 10^{-11}$

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Question

Would the molar solubility of Cr(OH)3 increase or decrease as the pH is lowered (i.e. made more acidic)?

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Answer

Since Cr(OH)3 is a basic salt, decreasing the pH makes it more soluble.

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Question

Calculate the molar solubility of SrF2 in 0.023M NaF. The Ksp for SrF2 is 4.3 x 10-9.

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Answer

$K_{sp} = [Sr^{2+} ] [F^- ]^2 = 4.3 x 10^{-9}$

$4.3 x 10^{-9} = [x] [0.023+2x]^2 \sim [x][0.023]^2$

$x = 8.1 x 10^{-6}$

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Question

Calculate the molar solubility of Mn(OH)2 at pH 9.5. The Ksp for Mn(OH)2 is 1.6 x 10-13.

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Answer

$Mn(OH)_2 \rightarrow Mn^{2+} + 2 : OH^-$

$K_{sp} = 1.6 x 10^{-13} = [Mn^{2+} ] [OH^- ]^2$

$[OH^- ] = 10^{-pOH} = 10^{-4.5} = 3.16 x 10^{-5}$

$K_{sp}= 1.6 x 10^{-13} = [x] [3.16 x 10^{-5} + 2x]^2 \sim [x] [3.16 x 10^{-5} ]^2$

$x = 1.6 x 10^{-4}$

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Question

Calculate the molar solubility of CaF2 (Ksp = 3.9 x 10-11) in a room temperature solution of 0.010 M Ca(C2H3O2)2.

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Answer

$CaF_2 \rightarrow Ca^{2+ } + 2 F^-$

$K_{sp} = 3.9 x 10^{-11} = [Ca^{2+} ] [F^- ]^2$

$K_{sp}= 3.9 x 10^{-11} = [x+0.01] [2x]^2 \sim [0.01] [2x]^2$

$x = 3.1 x 10^{-5}$

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Question

Calculate the molar solubility of AgBr in 0.050 M AgNO3 at room temperature. The Ksp of AgBr is 5.4 x 10-13.

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Answer

$K_sp = [Ag^+ ][Br^- ] = 5.4 x 10^{-13}$

$5.4 x 10^{-13} = [0.05+x][x] \sim [0.05][x]$

$x = 1.08 x 10^{-11}$

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Question

Would the molar solubility of Cr(OH)3 increase or decrease as the pH is lowered (i.e. made more acidic)?

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Answer

Since Cr(OH)3 is a basic salt, decreasing the pH makes it more soluble.

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Question

Calculate the molar solubility of SrF2 in 0.023M NaF. The Ksp for SrF2 is 4.3 x 10-9.

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Answer

$K_{sp} = [Sr^{2+} ] [F^- ]^2 = 4.3 x 10^{-9}$

$4.3 x 10^{-9} = [x] [0.023+2x]^2 \sim [x][0.023]^2$

$x = 8.1 x 10^{-6}$

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Question

Calculate the molar solubility of Mn(OH)2 at pH 9.5. The Ksp for Mn(OH)2 is 1.6 x 10-13.

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Answer

$Mn(OH)_2 \rightarrow Mn^{2+} + 2 : OH^-$

$K_{sp} = 1.6 x 10^{-13} = [Mn^{2+} ] [OH^- ]^2$

$[OH^- ] = 10^{-pOH} = 10^{-4.5} = 3.16 x 10^{-5}$

$K_{sp}= 1.6 x 10^{-13} = [x] [3.16 x 10^{-5} + 2x]^2 \sim [x] [3.16 x 10^{-5} ]^2$

$x = 1.6 x 10^{-4}$

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Question

Calculate the molar solubility of CaF2 (Ksp = 3.9 x 10-11) in a room temperature solution of 0.010 M Ca(C2H3O2)2.

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Answer

$CaF_2 \rightarrow Ca^{2+ } + 2 F^-$

$K_{sp} = 3.9 x 10^{-11} = [Ca^{2+} ] [F^- ]^2$

$K_{sp}= 3.9 x 10^{-11} = [x+0.01] [2x]^2 \sim [0.01] [2x]^2$

$x = 3.1 x 10^{-5}$

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Question

Calculate the molar solubility of AgBr in 0.050 M AgNO3 at room temperature. The Ksp of AgBr is 5.4 x 10-13.

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Answer

$K_sp = [Ag^+ ][Br^- ] = 5.4 x 10^{-13}$

$5.4 x 10^{-13} = [0.05+x][x] \sim [0.05][x]$

$x = 1.08 x 10^{-11}$

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Question

Would the molar solubility of Cr(OH)3 increase or decrease as the pH is lowered (i.e. made more acidic)?

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Answer

Since Cr(OH)3 is a basic salt, decreasing the pH makes it more soluble.

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Question

Calculate the molar solubility of SrF2 in 0.023M NaF. The Ksp for SrF2 is 4.3 x 10-9.

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Answer

$K_{sp} = [Sr^{2+} ] [F^- ]^2 = 4.3 x 10^{-9}$

$4.3 x 10^{-9} = [x] [0.023+2x]^2 \sim [x][0.023]^2$

$x = 8.1 x 10^{-6}$

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Question

Calculate the molar solubility of Mn(OH)2 at pH 9.5. The Ksp for Mn(OH)2 is 1.6 x 10-13.

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Answer

$Mn(OH)_2 \rightarrow Mn^{2+} + 2 : OH^-$

$K_{sp} = 1.6 x 10^{-13} = [Mn^{2+} ] [OH^- ]^2$

$[OH^- ] = 10^{-pOH} = 10^{-4.5} = 3.16 x 10^{-5}$

$K_{sp}= 1.6 x 10^{-13} = [x] [3.16 x 10^{-5} + 2x]^2 \sim [x] [3.16 x 10^{-5} ]^2$

$x = 1.6 x 10^{-4}$

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Question

Calculate the molar solubility of CaF2 (Ksp = 3.9 x 10-11) in a room temperature solution of 0.010 M Ca(C2H3O2)2.

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Answer

$CaF_2 \rightarrow Ca^{2+ } + 2 F^-$

$K_{sp} = 3.9 x 10^{-11} = [Ca^{2+} ] [F^- ]^2$

$K_{sp}= 3.9 x 10^{-11} = [x+0.01] [2x]^2 \sim [0.01] [2x]^2$

$x = 3.1 x 10^{-5}$

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Question

Would the molar solubility of Cr(OH)3 increase or decrease as the pH is lowered (i.e. made more acidic)?

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Answer

Since Cr(OH)3 is a basic salt, decreasing the pH makes it more soluble.

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Question

Calculate the molar solubility of AgBr in 0.050 M AgNO3 at room temperature. The Ksp of AgBr is 5.4 x 10-13.

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Answer

$K_sp = [Ag^+ ][Br^- ] = 5.4 x 10^{-13}$

$5.4 x 10^{-13} = [0.05+x][x] \sim [0.05][x]$

$x = 1.08 x 10^{-11}$

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Question

Calculate the molar solubility of SrF2 in 0.023M NaF. The Ksp for SrF2 is 4.3 x 10-9.

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Answer

$K_{sp} = [Sr^{2+} ] [F^- ]^2 = 4.3 x 10^{-9}$

$4.3 x 10^{-9} = [x] [0.023+2x]^2 \sim [x][0.023]^2$

$x = 8.1 x 10^{-6}$

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Question

Calculate the molar solubility of Mn(OH)2 at pH 9.5. The Ksp for Mn(OH)2 is 1.6 x 10-13.

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Answer

$Mn(OH)_2 \rightarrow Mn^{2+} + 2 : OH^-$

$K_{sp} = 1.6 x 10^{-13} = [Mn^{2+} ] [OH^- ]^2$

$[OH^- ] = 10^{-pOH} = 10^{-4.5} = 3.16 x 10^{-5}$

$K_{sp}= 1.6 x 10^{-13} = [x] [3.16 x 10^{-5} + 2x]^2 \sim [x] [3.16 x 10^{-5} ]^2$

$x = 1.6 x 10^{-4}$

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Question

Calculate the molar solubility of CaF2 (Ksp = 3.9 x 10-11) in a room temperature solution of 0.010 M Ca(C2H3O2)2.

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Answer

$CaF_2 \rightarrow Ca^{2+ } + 2 F^-$

$K_{sp} = 3.9 x 10^{-11} = [Ca^{2+} ] [F^- ]^2$

$K_{sp}= 3.9 x 10^{-11} = [x+0.01] [2x]^2 \sim [0.01] [2x]^2$

$x = 3.1 x 10^{-5}$

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