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AP Chemistry : Solubility and Equilibrium

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Example Questions

Example Question #1 : Solubility And Equilibrium

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

Possible Answers:

1.08 M

2.16 M

1.08 x 10^-11 M

1.57 x 10^-12 M

2.16 x 10^-11 M

Correct answer:

1.08 x 10^-11 M

Explanation:

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

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

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

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Example Question #1 : Solubility And Equilibrium

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

Possible Answers:

Increase

There is no change

Decrease

Can not be determined

Correct answer:

Increase

Explanation:

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

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Example Question #2 : Solubility And Equilibrium

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

Possible Answers:

3.2 x 10^-3 M

3.2 x 10^-5 M

8.1 x 10^-6 M

5.2 x 10^-4 M

1.6 x 10^-3 M

Correct answer:

8.1 x 10^-6 M

Explanation:

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

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

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

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Example Question #41 : Solutions

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

Possible Answers:

1.6 x 10^-4 M

3.5 x 10^-4 M

2.4 x 10^-3 M

1.5 x 10^-3 M

2.1 x 10^-5M

Correct answer:

1.6 x 10^-4 M

Explanation:

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

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

pOH = 14 - pH = 14 - 9.5 = 4.5

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

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

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

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Example Question #42 : Solutions

Calculate the molar solubility of CaF₂(K_sp = 3.9 x 10^-11) in a room temperature solution of 0.010 M Ca(C₂H₃O₂)_2.

Possible Answers:

3.2 x 10^-2 M

4.2 x 10^-4M

3.7 x 10^-4M

3.1 x 10^-5 M

1.6 x 10^-2 M

Correct answer:

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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AP Chemistry : Solubility and Equilibrium

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #1 : Solubility And Equilibrium

Example Question #1 : Solubility And Equilibrium

Example Question #2 : Solubility And Equilibrium

Example Question #41 : Solutions

Example Question #42 : Solutions

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