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AP Chemistry : Acid-Base Reactions

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Example Questions

Example Question #11 : Reaction Types

A chemist boils off the water from 627 mL of a solution with a pH of 10.2. The only dissolved compound in the solution is $NaOH$ . How much solid remains?

Possible Answers:

$2.31\times 10^{-3}\hspace{1 mm}g$

$1.01\times 10^{-4}\hspace{1 mm}g$

$1.71\times 10^{-3}\hspace{1 mm}g$

No solid remains.

$4.03\times 10^{-3}\hspace{1 mm}g$

Correct answer:

$4.03\times 10^{-3}\hspace{1 mm}g$

Explanation:

The pH of the solution is 10.2, and we know:

$pH\hspace{1 mm}+\hspace{1 mm}pOH=14$

$pOH=14-pH=14-10.2=3.8$

$pOH=-log[OH^-]$

$3.8=-log[OH^-]$

$-3.8=log[OH^-]$

$[OH^-]=10^{-3.8}=1.58\times 10^{-4}$

And since $NaOH_{(s)}\rightarrow Na^+_{(aq)}+OH^-_{(aq)}$ , $[NaOH]=1.58\times 10^{-4}$ .

$627\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{1.58\times 10^{-4}\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{40\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=4.03\times 10^{-3}\hspace{1 mm}g\hspace{1 mm}NaOH$

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Example Question #12 : Reactions And Equilibrium

A chemist adds 10mL of a solution with a pH of 3.00 to 50mL of water. What is the pH of the new solution?

Possible Answers:

3.012

3.78

4.25

3.52

2.22

Correct answer:

3.78

Explanation:

First, we will determine the amount of hydronium ions in the first solution. This allows us to measure the moles of hydronium ion that are transferred to water.

pH=-log[H^+]

3.00=-log[H^+]

$[H+]=10^{-3}=0.00100\hspace{1 mm}M$

$10\hspace{1 mm}mL\times\frac{1\hspace{ 1 mm}L}{1000\hspace{ 1 mm}mL}\times\frac{0.001\hspace{ 1 mm}mol}{1\hspace{ 1 mm}L}=1.00\times10^{-5}\hspace{ 1 mm}mol$

Now we can calculate the hydronium ion concentration in the new solution, using the amount transferred and the volume of water, and can solve for pH.

$\frac{1.0\times10^{-5}mol}{60\hspace{ 1 mm}mL}\times\frac{1000\hspace{ 1 mm}mL}{1\hspace{ 1 mm}L}=1.67\times 10^{-4}\hspace{ 1 mm}M$

$pH=-log[1.67\times 10^{-4}]=3.78$

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Example Question #12 : Reactions And Equilibrium

Which two solutions, when combined in equal amounts, will produce the lowest pH?

Possible Answers:

0.5M acetic acid and 1.0M NaCl

2.5M NH₃ and 2.5M NH₄⁺

1.0M HCl and 1.0M NaCl

2.0M HCl and 2.0M KOH

Correct answer:

1.0M HCl and 1.0M NaCl

Explanation:

HCl is a strong acid, while NaCl is a salt. Mixing these two will result in a pH < 1.

As for the other choices, mixing 2.0M HCl and 2.0M KOH will result in the formation of KCl and H₂O, resulting in a neutral pH. Acetic acid (pKa= 4.76) is a weak acid, so it will have a pH less than 7 but greater than 1. Finally, NH₃ and NH₄⁺ is a buffer system between a weak base and weak acid, and if equal amounts are present, the pH will be the same as the pKa of NH₄⁺, which is around 9.

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Example Question #14 : Reactions And Equilibrium

1mol of hydrobromic acid (HBr) is added to water, resulting in 1L of solution. Worried that the acid is too strong, a professor adds more water, increasing the volume of solution to 2L.

What was the pH of the solution before adding water? What was the pH after adding the water?

Possible Answers:

Before, pH = 0; after, pH = 0.3

Before, pH = 0; after, pH =-0.3

Before, pH = 1; after, pH = 0.5

Before, pH = 1; after, pH = 2

Correct answer:

Before, pH = 0; after, pH = 0.3

Explanation:

The pH of a solution is given by the equation pH=-log[H^+] .

1) We can assume that hydrobromic acid, as a strong acid, will dissociate completely. As a result, there will be 1mol of protons in the initial solution. Molarity is a method of measuring the concentration of an agent in a solution and is defined by the equation below.

$M=\frac{mol\ solute}{L\ solution}$

Since we have 1mol of protons and 1L of solution, the molarity of protons is simply 1M in the initial solution (1mol/1L). We can plug this into the pH equation and solve.

pH=-log[1]=0

2) After the dilution with water, the concentration of protons will change. Since the volume of the solution has doubled, the concentration of protons has been halved.

$M=\frac{1mol}{2L}=0.5M$

pH=-log[H^+]=-log[0.5]=0.3

* Key points to remember!

1) The lower the pH, the more acidic the solution. This is why the diluted solution was 0.3 and the original, more acidic solution was 0.

2) pH is a logarithmic scale. As a result, a pH of 2 is ten times more acidic than a pH of 3. This is why halving the concentration DID NOT halve the pH.

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Example Question #15 : Reactions And Equilibrium

What is the pH of a solution if you dilute 10mL of a 0.56M NaOH solution into 100mL of water?

Possible Answers:

12.75

1.25

12.71

There is not enough information to answer this question

1.29

Correct answer:

12.71

Explanation:

We first need to calculate the moles of NaOH in 10mL of a 0.56M solution.

$10\hspace{1 mm}mL\cdot\frac{0.56\hspace{1 mm}mol\hspace{1 mm}NaOH}{1000\hspace{1 mm}mL}=5.6\cdot 10^{-3}\hspace{1 mm}mol\hspace{1 mm} NaOH$

Now, in the new volume, the concentration is equal to this same number of moles divided by the new volume. The new volume is the sum of the two solutes together, so the sum of 10mL of solution and 100mL of water.

$\frac{5.6\cdot 10^{-3}\hspace{1 mm}mol\hspace{1 mm}NaOH}{110\hspace{1 mm}mL}\cdot\frac{1000\hspace{1 mm}mL}{1\hspace{1 mm}L}= 5.09\cdot 10^{-2}\hspace{1 mm}M$

Now we need to calculate pH. Since NaOH will dissociate one hydroxide ion for every molecule of NaOH in solution, the concentration of hydroxide is equal to the concentration of NaOH.

$[NaOH]=[OH^-]=5.09\cdot 10^{-2}M$

$pOH=-log[OH^-]=-log[5.09\cdot 10^{-2}]=1.29$

Knowing the pH allows us to solve for the pH.

pH+pOH=14

pH=14-pOH=14-1.29=12.71

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Example Question #16 : Reactions And Equilibrium

What is the pOH of a solution that has a [H⁺] = 3.2 X 10^-7 mol?

Possible Answers:

-7

6.5

7.5

3.2

Correct answer:

7.5

Explanation:

[H⁺] = 3.2 X 10^-7 mol

pH = -log [H⁺]

= 7 - log 3.2

= 7 -.505

=6.495

pOH = 14 - pH

= 7.5

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Example Question #13 : Reactions And Equilibrium

What is the range of possible hydrogen ion concentrations in basic solution?

Possible Answers:

10^–1 to 10^–6.999

10^–13 to 10^–7.001

2 to 3

7 to 14

Correct answer:

10^–13 to 10^–7.001

Explanation:

For a solution to be basic, it must have a pH between 8 and 14, since 7 is neutral; pH is –log(hydrogen ion concentration); the range of possible values is between 10^–13 and 10^–7.001.

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Example Question #21 : Reactions And Equilibrium

A solution has a hydrogen ion concentration of $2.5 * 10^{-6}$ . What is the pH of this solution?

Possible Answers:

6.5

5.6

7.2

4.7

5.0

Correct answer:

5.6

Explanation:

pH can be found by taking the negative log of the hydrogen ion concentration.

pH=-log[H^+]

$[H^+]=2.5*10^{-6}$

$pH = -log(2.5 *10^{-6})$
We may not know the exact value of this calculation, but we know that it will be somewhere between 5 and 6.

$10^{-5}>2.5*10^{-6}>10^{-6}$

$-log(10^-5)<-log(2.5*10^{-6})<-log(10^{-6})$

$5<-log(2.5*10^{-6})<6$

Only one answer choice fits this range. The answer is about 5.6.

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Example Question #14 : P H

Find the pH of a 0.03M solution of nitric acid.

Possible Answers:

1.52

0.03

3.0

12.47

Correct answer:

1.52

Explanation:

Nitric acid is a strong acid, meaning that it will dissociate completely in water.

$HNO_3\rightleftharpoons H^++NO_3^-$

As a result, a 0.03M nitric acid solution will dissociate completely, resulting in a proton concentration of 0.03M.

$0.30M\ NO_3*\frac{1mol\ H^+}{1mol\ HNO_3}=0.30M\ H^+$

Now that we know the proton concentration, we can use the equation for pH to find the pH value of the solution.

$\small pH = -log[H^{+}]$

$\small pH = -log[0.03]$

$\small pH = 1.52$

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Example Question #11 : Acid Base Reactions

Find the pH of a 0.1M solution of hydrogen cyanide.

$\small K_{a} = 6.2*10^{-10}$

Possible Answers:

4.63

5.10

1.01

8.96

Correct answer:

5.10

Explanation:

Hydrogen cyanide will dissociate according to the following reaction.

$\small HCN_{(aq)} \rightarrow H^{+}_{(aq)} + CN^{-}_{(aq)}$

Hydrogen cyanide is a weak acid, meaning that it will not dissociate completely in solution. We need to use the acid dissociation constant in order to determine the concentration of protons in solution at equilibrium. We can solve for this value using an ICE table.

I. The initial concentration for the acid of 0.1M. There are no protons or cyanide ions in the solution initially, so they both have concentrations of 0.0M.

C. The concentration of both ions will increase by an unknown amount, $\small x$ , while the acid concentration will decrease by a concentration of $\small x$ .

E. Write the equilibrium expression and compare it to the dissociation constant.

$\small K_{a} = \frac{[H^{+}][CN^{-}]}{[HCN]} = 6.2*10^{-10}$

$\small \frac{[x][x]}{[0.1-x]} = 6.2*10^{-10}$

The value of $\small x$ will be small enough that it is negligible in comparison to 0.1, and can be ignored in the denominator.

$\frac{[x][x]}{0.1}=6.2*10^{-10}$

$x^2=6.2*10^{-11}$

$\small x = 7.87*10^{-6}$

From our set up, we know that the value of $\small x$ is equivalent to the concentration of hydrogen ions at equilibrium. Using this value, we can calculate the pH of the solution.

$pH=-\log{[H^+]}$

$pH=-\log{(7.87*10^{-6})}$

pH=5.10

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AP Chemistry : Acid-Base Reactions

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #11 : Reaction Types

Example Question #12 : Reactions And Equilibrium

Example Question #12 : Reactions And Equilibrium

Example Question #14 : Reactions And Equilibrium

Example Question #15 : Reactions And Equilibrium

Example Question #16 : Reactions And Equilibrium

Example Question #13 : Reactions And Equilibrium

Example Question #21 : Reactions And Equilibrium

Example Question #14 : P H

Example Question #11 : Acid Base Reactions

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