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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 #1 : 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^-5 M

8.1 x 10^-6 M

5.2 x 10^-4 M

1.6 x 10^-3 M

3.2 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:

2.1 x 10^-5M

1.5 x 10^-3 M

1.6 x 10^-4 M

2.4 x 10^-3 M

3.5 x 10^-4 M

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:

1.6 x 10^-2 M

3.2 x 10^-2 M

3.1 x 10^-5 M

4.2 x 10^-4M

3.7 x 10^-4M

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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Example Question #1 : Other Solution Concepts

A solution on NaCl has a denisty of 1.075 g/mL. If there are 0.475 L of solution present, what is the mass?

Possible Answers:

2.263 g

441.9 g

0.510 g

510.6 g

2.2 g

Correct answer: 510.6 g

Explanation:

1.075 g / mL * 0.475 L

First, convert to mL

1.075 g / mL * 475 mL = 510.6 g

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Example Question #43 : Solutions

5L of 0.1M NaCl and 10L of 0.2M NaI are combined in a single vessel.

What is the final concentration of sodium ions in solution?

Possible Answers:

0.1M

0.20M

0.2M

0.17M

Correct answer:

0.17M

Explanation:

NaCl and NaI are both highly soluble; thus both solutions can be treated as containing separate ions of sodium, chloride, and iodide.The final concentration can be found by finding the total number of moles of sodium ions and the total volume from both solutions.

$M=\frac{\text{total moles}\ Na^+}{\text{total volume}}$

We can find the moles of sodium ions from each solution by multiplying the volume by the molar concentration.

$NaCl:\ 5L*\frac{0.1mol\ NaCl}{1L}*\frac{1mol\ Na^+}{1mol\ NaCl}=0.5mol\ Na^+$

$NaI:\ 10L*\frac{0.2mol\ NaI}{1L}*\frac{1mol\ Na^+}{1mol\ NaI}=2mol\ Na^+$

The total moles of sodium ions is:

$2mol+0.5mol=2.5mol\ Na^+$

Divide this by the total solution volume to find the final concentration:

$M=\frac{2.5mol\ Na^+}{5L+10L}=0.17M$

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Example Question #1 : Other Solution Concepts

Given that the pKa of acetic acid is 4.76, what is the percentage of the protonated form of acetic acid in a solution where the pH is 6?

Possible Answers:

$5.44\%$

$66.6\%$

$94.56\%$

$33.3\%$

There is not enough information given

Correct answer:

$5.44\%$

Explanation:

In order to solve this problem, we first must use the Henderson-Hasselbalch equation:

$pH=pKa+log\left [ \frac{conjugate\: base}{acid} \right ]$

Since we are given the pH of the solution and the pKa of acetic acid, we are able to solve for the ratio of conjugate base to acid:

$6=4.76+log\left [ \frac{conjugate\: base}{acid} \right ]$

$1.24=log\left [ \frac{conjugate\: base}{acid} \right ]$

$10^{1.24}=\left [\frac{conjugate\: base}{acid}\right ]=17.38$

Now that we have the ratio of conjugate base to acid, we need to calculate the percentage of the acid, or protonated form, in solution. To do this, it's important to realize that for every 17.38 moles of conjugate base, there is 1 mol of acid. Therefore, the total amount of acetic acid + acetate is equal to 18.38.

$Percent\:protonated\:form=\frac{acetic\: acid}{acetic\: acid+acetate}\cdot 100\%$

$\frac{acetic\: acid}{acetic\: acid+acetate}=\frac{1}{1+17.38}=\frac{1}{18.38}$

$Percent\:protonated\:form=\frac{1}{18.38}\cdot 100\%=5.44\%$

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Example Question #4 : Other Solution Concepts

Suppose that two containers, and , contain equal amounts of water. If 5 moles of NaCl is added to solution and 5 moles of glucose is added to solution , which solution will experience a greater increase in boiling point?

Possible Answers:

There is not enough information to answer the question

Solution , because NaCl is able to dissociate into $Na^{+}$ and $Cl^{-}$ ions, thus resulting in a greater amount of particles dissolved in solution

Solution , because glucose has a greater molar mass than NaCl

Neither solution will experience a change in boiling point

Both solutions will exhibit the same change in boiling point

Correct answer:

Solution , because NaCl is able to dissociate into $Na^{+}$ and $Cl^{-}$ ions, thus resulting in a greater amount of particles dissolved in solution

Explanation:

In the question stem, we are told that equal molar amounts of NaCl and glucose are added to containers and , respectively. The change in boiling point of water is a colligative property that is dependent on the number of dissolved solute particles, regardless of their identity. The addition of 5 moles of NaCl will result in approximately 10 moles of dissolved solute, since each mol of NaCl can dissociate into two ions, according to the following reaction:

$NaCl(s)\rightarrow Na^{+}\left( aq\right )+Cl^{-}\left( aq\right )$

Glucose, on the other hand, does not dissociate and simply remains as intact molecules. Thus, the addition of 5 moles of glucose to container results in 5 moles of dissolved solute. Since solution contains approximately twice as many dissolved solute particles as does solution , it will experience a greater increase in the boiling point of water.

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Example Question #51 : Solutions

Approximately what is the pH of a $3.7\cdot 10^{-4}M$ solution of NaOH at $25^{o}C$ ?

Possible Answers:

10.57

9.15

12.18

There is insufficient information to answer the question

3.43

Correct answer:

10.57

Explanation:

We are given the concentration of NaOH in solution and asked to find the pH. To do this, we must make use of the following equation:

$pOH=-log\left [ OH^{-}\right ]$

It is also important to realize that NaOH is a strong base and will thus dissociate completey according to the following reaction:

$NaOH_{(s)} \rightarrow Na^{+}_{(aq)}+OH^{-}_{(aq)}$

Thus, for every one mol of NaOH that reacts, an equal number of moles of $OH^{-}$ will be produced. And since there are $3.7\cdot 10^{-4}moles$ NaOH to begin with, then $3.7\cdot 10^{-4}moles$ $OH^{-}$ will be produced.

$pOH=log(3.7\cdot 10^{-4}M)=3.43$

Remember that this calculated value so far is the pOH, not the pH! To calculate the pH, it is vital to remember that pH+pOH=14 at $25^{o}C$ . Thus,

pH=14-3.43=10.57

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AP Chemistry : AP Chemistry

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 #1 : Solubility And Equilibrium

Example Question #41 : Solutions

Example Question #42 : Solutions

Example Question #1 : Other Solution Concepts

Example Question #43 : Solutions

Example Question #1 : Other Solution Concepts

Example Question #4 : Other Solution Concepts

Example Question #51 : Solutions

All AP Chemistry Resources

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