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MCAT Physical : Acid-Base Equilibrium

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Example Questions

Example Question #11 : Reactions And Titrations

A solution of hydrofluoric acid has a concentration of 0.3M

The $K_{a}$ for is $7.2*10^{-4}$ .

What is the pH of the solution?

Possible Answers:

2.58

3.33

5.46

1.83

Correct answer:

1.83

Explanation:

Since hydrofluoric acid is a weak acid, an ICE table needs to be set up in order to determine the hydronium ion concentration. Since both fluoride ion and hydronium ion concentrations will increase by , while the acid concentration will decrease by , the equilibrium expression comes out to be:

$K_a=\frac{[H^+][F^-]}{[HF]}$

$7.2*10^{-4}=\frac{x^{2}}{0.3-x}$

Note that the in the denominator will have a negligible effect and can be ignored.

$7.2*10^{-4}=\frac{x^{2}}{0.3}$

x = 0.015

Since is equal to the hydronium ion concentration, we can calculate the pH by taking the negative log of the concentration:

-log[0.015] = 1.83

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Example Question #12 : Acid Base Chemistry

A solution of hydrofluoric acid has a concentration of 0.3M .

The $K_{a}$ for is $7.2*10^{-4}$ .

If sodium hydroxide is slowly added to the acid solution, what will the pH be at the equivalence point?

Assume the concentration of the acid is not changed with the addition.

Possible Answers:

9.45

7.00

5.69

8.31

Correct answer:

8.31

Explanation:

At the equivalence point, there are equimolar amounts of acid and base. This means that all weak acid has been neutralized, and only the conjugate base remains. Since the conjugate base of a weak acid will affect the pH, we need to use an ICE table in order to find the pH. First, we start by finding the base dissociation constant of the conjugate base, using the equation:

$K_{a}K_{b} = K_{w}$

$(7.2*10^{-4})K_{b} = 1*10^{-14}$

$K_{b} = 1.39*10^{-11}$

The balanced equation for the conjugate base dissociation is:

$F^{-} + H_{2}O \rightleftharpoons HF + OH^{-}$

As the hydroxide ion and acid concentrations increase by , the fluoride ion concentration will decrease by . This makes the equilibrium expression:

$\frac{x^{2}}{0.3-x} = 1.39*10^{-11}$

$x = 2.04*10^{-6}$

Since this is the hydroxide concentration, we can find the pH by taking the negative log of this value, then subtracting from 14:

$pH = 14 - (-log(2.04*10^{-6})) = 14-5.69 = 8.31$

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Example Question #12 : Acid Base Equilibrium

$H_{2}O_{(l)}\ \leftrightharpoons \ H_{3}O^{+}_{(aq)}\ + OH^{-}_{(aq)}$

$K_{w} = 1 \cdot 10^{-14}$

Based on the above information, it is expected that __________.

Possible Answers:

The concentration of the cation is $1\cdot 10^{-7}M$

The concentration of the hydroxide ion is $1\cdot 10^{-14}M$

The product of the anion and cation concentrations is $1\cdot10^{-7}M^{2}$

There are an equal number of water molecules, hydronium ions, and hydroxide ions

Correct answer:

The concentration of the cation is $1\cdot 10^{-7}M$

Explanation:

$K_{w} = [H_{3}O^{+}] \cdot [OH^{-}] = 1 \cdot 10^{-14}$

Since the product of the cation and anion is $1\cdot10^{-14}$ , the only true statement is that the concentration of the cation $(H_{3}O^{+})$ is the square root of this number:

$\sqrt{1 \cdot 10^{-14}} =1 \cdot 10^{-7} M$

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MCAT Physical : Acid-Base Equilibrium

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #11 : Reactions And Titrations

Example Question #12 : Acid Base Chemistry

Example Question #12 : Acid Base Equilibrium

All MCAT Physical Resources

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