Solubility and Equilibrium

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AP Chemistry › Solubility and Equilibrium

Questions 1 - 5

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

1.08 x 10-11 M

2.16 x 10-11 M

1.08 M

2.16 M

1.57 x 10-12 M

Explanation

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

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

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

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

Increase

Decrease

There is no change

Can not be determined

Explanation

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

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

3.2 x 10-3 M

1.6 x 10-3 M

8.1 x 10-6 M

3.2 x 10-5 M

5.2 x 10-4 M

Explanation

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

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

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

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

1.5 x 10-3 M

2.4 x 10-3 M

3.5 x 10-4 M

1.6 x 10-4 M

2.1 x 10-5 M

Explanation

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

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

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

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

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

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

1.6 x 10-2 M

3.2 x 10-2 M

4.2 x 10-4M

3.7 x 10-4 M

3.1 x 10-5 M

Explanation

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

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

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

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

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