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AP Chemistry : pH

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Example Questions

Example Question #24 : Reactions And Equilibrium

0.01M sodium cyanide solution is added to a 2.0M solution of hydrogen cyanide.

The acid dissociation constant for HCN is $\small 6.2*10^{-10}$ .

Find the pH of the solution.

Possible Answers:

6.9

5.7

4.6

4.5

Correct answer:

6.9

Explanation:

Hydrogen cyanide will dissociate in solution based on the following reaction.

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

By adding sodium cyanide to the solution, we have increased the concentration of the right side of the reaction. Remembering Le Chatelier's principle, we would predict that this would push the reaction to the left, and decrease the dissociation of the acid. Sodium cyanide will dissolve completely in an aqueous solution, resulting in an initial 0.01M cyanide ion solution. Knowing this, we can set up an ICE table, and solve for the pH of the solution.

I. There is initially a 2.0M concentration of acid, and 0.01M concentration of cyanide ions. There are no protons initially in the solution.

C. Both ions will increase in concentration by a value of $\small x$ , while the acid's concentration will decrease by a value of $\small x$ .

E. Placing these values in the equilibrium expression, we can calculate the value of $\small x$ .

$\small 6.2*10^{-10} = \frac{[x][0.01+x]}{[2-x]}$

Since the value for $\small x$ will be much less than 0.01 or 2, we can omit the variable from these factors in the equation.

$\small 6.2*10^{-10} = \frac{[x][0.01]}{[2]}$

$x=1.24*10^{-7}M$

Based on our set up, this concentration is equal to the concentration of protons in solution.

$[H^+]=1.24*10^{-7}M$

Now we can use the equation for pH and the proton concentration to find the final pH of the solution.

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

$pH=-\log(1.24*10^{-7})=6.9$

Using the pH equation, we determine that the pH of the solution is 6.9. The solution is only slightly acidic due to the initial addition of cyanide ions.

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Example Question #21 : P H

Wine has an approximate pH of 4.0. How much more acidic is wine compared to water, which has a pH of 7.0?

Possible Answers:

3 times more acidic

1,000 times more acidic

30 times more acidic

100 times more acidic

Correct answer:

1,000 times more acidic

Explanation:

The pH scale is a logarithmic scale, meaning that every time the value decreases by one, there has been a ten-fold increase in the acidity of the solution. Since wine is three steps more acidic than neutral water, we can say that wine is 1000 times more acidic than water.

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

$7=-\log(10^{-7})$

$4=-\log(10^{-4})$

$10^{-4}=1000(10^{-7})$

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Example Question #26 : Reactions And Equilibrium

What is the pH of a solution of HCl with a concentration of 0.0002M?

Possible Answers:

7.2

3.7

2.0

3.1

1.7

Correct answer:

3.7

Explanation:

pH is a measurement of the hydrogen ion concentration in a solution:

pH=-log[H^+]

Hydrochloric acid is monoprotic and fully dissociates in solution; thus, the concentration of acid will be equal to the concentration of protons.

$0.0002M\ HCl*\frac{1mol\ H^+}{1mol\ HCl}=0.0002M\ H^+$

Now we can use the pH equation to calculate the pH form the proton concentration.

pH=-log(0.0002M)

pH=3.7

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Example Question #22 : P H

$\small 50.0mL$ of a $\small 1.5M$ solution of hydrochloric acid is treated with $\small 20.0mL$ of a $\small 2.0M$ solution of sodium hydroxide. What is the pH of the resulting solution?

Possible Answers:

1.60

3.30

5.70

0.30

8.20

Correct answer:

0.30

Explanation:

To find the pH, we must first consider how many moles of $\small H^ +$ are present initially, and how many moles of $\small OH^-$ are added. All initial protons will come from the acid, while all hydroxide ions will come from the base.

$0.05L\ HCl*\frac{1.5mol}{1L}*\frac{1mol\ H^+}{1mol\ HCl}=0.075mol\ H^+$

$0.02L\ NaOH*\frac{2.0mol}{1L}*\frac{1mol\ OH^-}{1mol NaOH}=0.040mol\ OH^-$

These are our starting amounts, before the reaction occurs.

$HCl+NaOH\rightarrow NaCl+H_2O$

The reaction occurs with a one-to-one ratio, meaning that the ion present in the less amount will be the limiting reagent. In this question, the hydroxide ions are the limiting reagent. We need to determine how many moles of protons remain after all of the sodium hydroxide has reacted.

$0.075mol\ H^+-0.040mol\ OH^-=0.035mol\ H^+$

The total volume of the final solution wil be equal to the sum of the two original solutions.

V=0.05L+0.02L=0.07L

We can now find the final concentration of protons after the reaction.

$\frac{0.035mol\ H^+}{0.07L}=0.5M\ H^+$

Use the formula for pH to find the final pH of the solution from the proton concentration.

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

$pH=-\log(0.5M)=0.30$

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Example Question #22 : P H

Sodium hydroxide is a strong base. What is the pH of a 0.02M sodium hydroxide solution?

Possible Answers:

1.7

2.0

12.0

12.3

Correct answer:

12.3

Explanation:

Since sodium hydroxide is a strong base, it will dissociate completely in water. This means that the concentration of the base will be equal to the concentration of hydroxide ions after the reaction runs to completion.

$NaOH\rightleftharpoons Na^++OH^-$

We can find the concentration of hydroxide ions via stoichiometry. One hydroxide ion is created from each molecule of sodium hydroxide that dissociates.

$0.02M\ NaOH*\frac{1mol\ OH^-}{1mol\ NaOH}=0.02M\ OH^-$

Since we have the concentration of hydroxide ions, we can solve for the pOH of the solution.

$\small pOH = -\log[OH^{-}]=-\log(0.02)$

$\small pOH = 1.70$

The question asks us to find the pH of the solution, so we will need to convert pOH to pH. To do so, we simply subtract the pOH from 14.

$pH+pOH=14\rightarrow pH=14-pOH$

$\small 14 - 1.70 = 12.3$

The pH of the solution is 12.3. Because sodium hydroxide is a strong base, it makes sense that the pH is above 7.

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Example Question #28 : Reactions And Equilibrium

Find the pH of a 0.1M solution of the weak base dimethylamine.

$\small K_{b} = 5.9*10^{-4}$

Possible Answers:

1.6

11.9

2.1

12.4

Correct answer:

11.9

Explanation:

Since dimethylamine is a weak base, we must compare the base dissociation constant to the equilibrium expression of the reaction. The reaction is written as follows:

$\small \small B_{(aq)} \rightarrow OH^{-}_{(aq)} + HB^{+}_{(aq)}$

We can determine the values for the equilibrium expression by using an ICE table:

I. There is initially a concentration of 0.1M of dimethylamine. Neither of the ions are initially present in the solution, giving them initial concentrations of 0.0M.

C. Both of the ions (products) will increase in concentration by an unknown amount, $\small x$ , while the base (reactant) will decrease in concentration by the value $\small x$ .

E. We can use the unknown concentration changes to formulate an equation for the equilibrium conditions.

$\small K_{b} = \frac{[OH^{-}][HB^{+}]}{[B]} = 5.9*10^{-4}$

$\small 5.9*10^{-4} = \frac{[x][x]}{[0.1-x]}$

The value of $\small x$ will be extremely small, making a negligible impact on the 0.1M denominator. The variable can be ignored in the denominator.

$\small 5.9*10^{-4} = \frac{[x][x]}{[0.1]}$

$5.9*10^{-5}=x^2$

$\small x = 7.7*10^{-3}M$

From our initial set up, we know that the value of $\small x$ is equivalent to the concentration of hydroxide ions in solution.

$[OH^-]=7.7*10^{-3}M$

Keep in mind that we are dealing with the dissociation of a base in this question. As a result, we can expect a pH above 7 since the overall solution will be basic. We can find the pOH of the solution by using the concentration of hydroxide ions.

$pOH=-\log[OH^-]=-\log(7.7*10^{-3}M)$

pOH=2.1

Now that we know the pOH, we can find the pH by using the following equation.

$pH+pOH=14\rightarrow 14-pOH=pH$

$\small 14- 2.1 = 11.9$

The pH of the solution is 11.9.

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Example Question #22 : P H

A chemist mixes 50.0mL of a $1.0M\ NaOH$ solution with 50.0mL of a $1.0M\ Ba(OH)_2$ solution. Assuming the two solutions are additive, what is the pH of the resulting solution?

Possible Answers:

12.17

10.90

14.18

13.82

12.75

Correct answer:

14.18

Explanation:

To find the pH of the solution after mixing the individual components, one first has to determine the final hydroxide concentration, [^-OH].

$[OH^-]=\frac{mol OH^-}{total volume}$

mol OH^-=(0.05L)(1.00M)+2(0.05L)(1.00M)=0.150mol (Remember that 1 mole of Ba(OH)₂ will give 2 moles of [^-OH]).

total volume=0.05L+0.05L=0.10L

$[OH^-]=\frac{0.150mol}{0.10L}=1.5M$

pOH=-log[OH^-]=-log(1.5M)=-0.18

pH=14-pOH=14-(-0.18)=14.18

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Example Question #61 : Ap Chemistry

A chemist adds 625g of solid $NaOH$ to 500mL of 16M $H_2SO_4$ . What is the pH of the solution after it reaches equilibrium?

Possible Answers:

2.1

0.76

1.4

1.8

0.12

Correct answer:

0.12

Explanation:

Consider the reaction of $NaOH$ and $H_2SO_4$ :

$2NaOH+H_2SO_4\rightarrow Na_2SO_4\hspace{1 mm}+2H_2O$

Now we will calculate the moles of $H_2SO_4$ in the solution prior to adding base.

$500\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{16\hspace{1 mm}moles\hspace{1 mm}H_2SO_4}{1\hspace{1 mm}L}=8.00\hspace{1 mm}moles\hspace{1 mm}H_2SO_4$

We will then calculate the amount of moles of $H_2SO_4$ that react with the base.

$625\hspace{1 mm}g\hspace{1 mm}NaOH\times\frac{1\hspace{1 mm}mole\hspace{1 mm}NaOH}{40\hspace{1 mm}g\hspace{1 mm}NaOH}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}H_2SO_4}{2\hspace{1 mm}moles\hspace{1 mm}NaOH}=7.81\hspace{1 mm}moles\hspace{1 mm}H_2SO_4$

We will then calculate the remaining moles of $H_2SO_4$ :

$8.00\hspace{1 mm}moles\hspace{1 mm}H_2SO_4 -\hspace{1 mm}7.81\hspace{1 mm}moles\hspace{1 mm}H_2SO_4=0.19\hspace{1 mm}moles\hspace{1 mm}H_2SO_4$

We will then calculate the new concentration of sulfuric acid:

$\frac{0.19\hspace{1 mm}moles\hspace{1 mm}H_2SO_4}{0.5\hspace{1 mm}L}=0.38\hspace{1 mm}M$

Sulfuric acid is a diprotic acid, so the hydrogen ion concentration is 0.76 M.

$pH=-log[H^{+}]=-log(0.76)=0.12$

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Example Question #23 : P H

If the Ksp of Mg(OH)2 is 1.2 * 10^–11 and the magnesium ion concentration is 1.2 * 10^–5M, at what pH does the Mg(OH)₂ compound begin to precipitate?

Possible Answers:

Correct answer:

Explanation:

The expression for Ksp is Ksp = [Mg₂₊][OH–]².

Thus, [OH–] = √(1.2 * 10^–11)/(1.2 * 10^–5)

[OH–] = 1 * 10^–3

Thus, pH = –log(1 X 10^–3) = 3

pH = 14 – 3 = 11

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Example Question #22 : Acid Base Reactions

The following is a list of acid dissociation constants for few acids.

Benzoic acid: $\small K_{a} = 6.3*10^{-5}$

Hydrofluoric acid: $\small K_{a} = 6.8*10^{-4}$

Nitrous acid: $\small K_{a} = 7.1*10^{-4}$

Propanoic acid: $\small K_{a} =1.3*10^{-5}$

One of these acids is used to create a 1M solution. It is determined that the solution has a pH of 2.44. Which of the following acids was used to make the solution?

Possible Answers:

Benzoic acid

Hydrofluoric acid

Nitrous acid

Propanoic acid

Correct answer:

Propanoic acid

Explanation:

Since the solution has a pH of 2.44, we can find the concentration of protons in the solution using the pH equation.

$pH=-\log[H^+]\rightarrow [H^+]=10^{-pH}$

$\small \small \small [H^{+}] = 10^{-2.44} = 3.6*10^{-3}M$

All of these acids are monoprotic acids, so the equilibrium expression for each acid follows the formula for dissociation.

$HA\rightarrow H^++A^-$

$\small K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}$

The concentration for protons will be equal to the concetration of the conjugate base, since they are both in a one-to-one ratio with the acid molcule.

$[H^+]=[A^-]=3.6*10^{-3}M$

Also, we can subtract the concentration of protons from the initial concentration of acid in order to find how much acid will remain at equilibrium.

$[HA]_{initial}-3.6*10^{-3}M=[HA]_{final}$

Using the terms for each concentration, we can try to solve the equilibrium expression.

$\small \small K_{a} = \frac{[3.6*10^{-3}][3.6*10^{-3}]}{[1-3.6*10^{-3}]}$

$K_a=\frac{1.3*10^{-5}}{0.9964}=1.3*10^{-5}$

Since the acid dissociation constant is equal to $\small 1.3*10^{-5}$ , we can look at the list and determine that propanoic acid was used to make the acidic solution.

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AP Chemistry : pH

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #24 : Reactions And Equilibrium

Example Question #21 : P H

Example Question #26 : Reactions And Equilibrium

Example Question #22 : P H

Example Question #22 : P H

Example Question #28 : Reactions And Equilibrium

Example Question #22 : P H

Example Question #61 : Ap Chemistry

Example Question #23 : P H

Example Question #22 : Acid Base Reactions

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