Solubility and Equilibrium - AP Chemistry

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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

$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^-$

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

$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^-$

$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

$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^-$

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

$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^-$

$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.

Tap to reveal answer

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)?

Tap to reveal answer

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

$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^-$

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

$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^-$

$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)?

Tap to reveal answer

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

$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^-$

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

$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^-$

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

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