MCAT Physical - Acid-Base Equilibrium

What is the pKa of acetic acid? (Ka = 1.8 * 10–5)

4.7

4.2

5.3

2.1

8.6

Explanation

We know that pKa is equal to –log(Ka). Thus, pKa of acetic acid is –log(1.8 * 10–5). This is not an easy problem to solve in your head, but there is a trick.

We know that 1 * 10–4 > 1.8 * 10–5 > 1 * 10–5, and we know that –log(1 * 10–4) = 4 and –log(1 * 10–5) = 5. Now we can conclude that our pKa is somewhere between 4 and 5.

Two answer choices fall in this range: 4.2 and 4.7.

1.8 * 10–5 is closer to 1 * 10–5 than it is to 1 * 10–4, so we can pick the answer closer to 5 than to 4 : 4.7.

What is the pH of a solution which has a hydroxide ion concentration of 5 * 10-4M?

$\small 10.70$

$\small 3.30$

$\small 2*10^{-11}$

$\small 5*10^{10}$

$\small 13.30$

Explanation

First convert concentration of OH- into concentration of H+. Remember that Kw is 1*10-14.

$\small [H^{+}]=\frac{1*10^{-14}}{[OH^{-}]} = \frac{1*10^{-14}}{5*10^{-4}}=2*10^{-11}$

Then, convert concentration of H+ into pH.

$\small pH = -log[H^{+}]=-log(2*10^{-11})\approx10.70$

Alternatively, you can use the hydroxide concentration to solve for pOH and convert to pH.

Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.

$HA \leftrightharpoons H^{+} + A^{-}$

All of the bases proceed in a similar fashion.

$BOH \leftrightharpoons B^{+} + OH^{-}$

The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.

Which of the following expressions most closely approximates the equilibrium constant of pure water?

$K_{w} = [}H^{+}}][OH^{-}]$

$K_w=\frac{[H^+]}{[OH^-]}$

$K_{w} = [}H^{+}}][H_{2}O]$

$K_{w} = [H_{2}O][OH^{-}]$

$K_{w} = [}H^{+}}]$

Explanation

The autionization of water proceeds as follows:

$H_{2}O \leftrightharpoons H^{+} + OH^{-}$

The equilibrium constant is calculated by the following formula.

$K_{eq}=\frac{[product]}{[reactant]}$

The equilibrium expression is $K_{w} = [}H^{+}}][OH^{-}]$ , with the pure water reactant excluded. Pure solids and liquids are not included in the equilibrium calculation.

What is the hydroxide concentration in an aqueous solution with a pH of 2?

$1*10^{-12}M$

There are no hydroxide ions in the solution

$1*10^{-2}M$

$1*10^{-4}M$

Explanation

The hydroxide concentration can be determined by considering the autoionization of water in a solution. At , water has the equilibrium constant of $K_{w} = 1*10^{-14}$ . This value is based on the autoionization of water.

$H_2O\rightleftharpoons H^++OH^-$

$K_w=[H^+][OH^-]=1*10^{-14}$

Since the reaction is in terms of proton and hydroxide ion concentrations, we can set the expression equal to this value in order to determine the concentration of hydroxide ions at the given pH. We start by determining the concentration of protons in the solution by using the following equation:

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

$[H^{+}] = 10^{-2} = 1*10^{-2}M$

Now, we can solve for the hydroxide concentration.

$K_w=[H^+][OH^-]=1*10^{-14}$

$[OH^{-}][1*10^{-2}] = 1*10^{-14}$

$[OH^{-}] = 1*10^{-12}M$

This problem can also be solved by calculating the pOH of the solution and using this value to find the hydroxide ion concentration.

A solution of hydrofluoric acid has a concentration of

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

What is the pH of the solution?

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.

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

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

$HA \leftrightharpoons H^{+} + A^{-}$

All of the bases proceed in a similar fashion.

$BOH \leftrightharpoons B^{+} + OH^{-}$

In aqueous conditions the equilibrium constant for a Brønsted-Lowry base, $A^{-}$ , can be expressed as which of the following?

$K_b=\frac{[HA][OH^-]}{[A^-]}$

$K_b=\frac{[HA][OH^-]}{[H_2O]}$

$K_b=\frac{[A^-][OH^-]}{[HA]}$

$K_b=\frac{[HA]}{[A^-]}$

$K_b=\frac{[OH^-]}{[A^-]}$

Explanation

The base reaction will essentially be the reverse of the acid reaction. In the question, the aqueous conditions mean that the base, $A^{-}$ , reacts with water to give the following reaction:

$A^{-} +H_{2}O \leftrightharpoons HA + OH^{-}$

Following normal equilibrium convention, we omit water from the equation because it is a pure liquid.

$K_{eq}=\frac{[product]}{[reactant]}$

$K_b=\frac{[HA][OH^-]}{[A^-]}$

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

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

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

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

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

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

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:

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

As the value of Ka increases, __________

all of these are true

the strength of the acid increases

pKa decreases

the strength of the conjugate base decreases

the reaction HA → H+ + A– favors the products

Explanation

Ka represents the equilibrium constant for the acid dissociation in water, HA → H+ + A–, so it is a measure of the products divided by the reactants. As this value increases, the reaction favors the products (the ions) more, meaning that the acid dissociates more. The definition of a strong acid is one that fully dissociates in water, so as Ka increases the strength of the acid increases, and the strength of the conjugate base decreases. pKa is defined as the –log(Ka), so as Ka increases pKa decreases. All of these answer choices are correct.

HCN dissociates based on the following reaction.

$HCN + H_{2}O \rightarrow CN^{-}+ H_{3}O^{+}$

The Ka for hydrogen cyanide is $\small 6.2*10^{-10}$ .

What is the Kb for CN-?

$\small 1.6*10^{-5}$

$\small 6.2*10^{4}$

$\small 6.2*10^{-24}$

$\small 1.3*10^{-4}$

Explanation

Remember that the Ka for the acid and the Kb for the conjugate base, when multiplied will equal the autoionization of water constant (Kw).

$K_{a} K_{b} = K_{w}=\small 1*10^{-14}$

$\small K_{b} = \frac{K_w}{K_a}=\frac{1*10^{-14}}{6.2*10^{-10}}$

$\small K_{b}= 1.6*10^{-5}$

$H_{2}CO_{3} \left ( K_{a} = 4.5 * 10^{-7} \right )$

$HCHO_{2} \left ( K_{a} = 1.8 * 10^{-4} \right )$

$HC_{2}H_{3}O_{2} \left ( K_{a} = 1.8 * 10^{-5} \right )$

$HF \left ( K_{a} = 6.3 * 10^{-4} \right )$

Given the above values of Ka, place the acids in order from strongest to weakest.

$HF > HCHO_{2} > HC_{2}H_{3}O_{2} > H_{2}CO_{3}$

$H_{2}CO_{3} > HC_{2}H_{3}O_{2} > HCHO_{2} > HF$

$HCHO_{2} > HF > HC_{2}H_{3}O_{2} > H_{2}CO_{3}$

$HF > H_{2}CO_{3} > HC_{2}H_{3}O_{2} > HCHO_{2}$

None of the above.

Explanation

The acid dissociation constant, Ka, describes how strongly an acid tends to break apart into hydrogen ions (H+) and its conjugate base (A-). The higher the dissociation constant, the stronger the acid. HF has the largest Ka of these acids, making it the strongest, and H2CO3 has the smallest Ka, making it the weakest.

Acid-Base Equilibrium

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation