AP Chemistry 8 Acids and Bases Questions

Explore AP Chemistry Unit 8 questions on pH and pOH, acid-base equilibria, titration curves, buffers, conjugate species, pKa, and pH-dependent solubility.

Covered syllabus points

8.1 Introduction to Acids and Bases 8.2 pH and pOH of Strong Acids and Bases 8.3 Weak Acid and Base Equilibria 8.4 Acid-Base Reactions and Buffers 8.5 Acid-Base Titrations 8.6 Molecular Structure of Acids and Bases 8.7 pH and pKa 8.8 Properties of Buffers 8.9 Henderson-Hasselbalch Equation 8.10 Buffer Capacity 8.11 pH and Solubility

[Maximum number: 5]

A student is given 50.0 mL of a solution of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ of unknown concentration. To determine the concentration of the solution, the student mixes the solution with excess $1.0 M \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)$ , causing a precipitate to form. The balanced equation for the reaction is shown below.

\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q) \rightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{CaCO}_{3}(s)

(a)

The student decides to first determine [ $\mathrm{OH}^{-}$ ] in the solution, then use that result to calculate the initial concentration of $\mathrm{CO}_{3}{ }^{2-}(a q)$ .

[ 3 ]

(i)

Identify a laboratory method (not titration) that the student could use to collect data to determine $\left[\mathrm{OH}^{-}\right]$ in the solution.

[ 1 ]

(ii)

Explain how the student could use the measured value in part (f)(i) to calculate the initial concentration of $\mathrm{CO}_{3}{ }^{2-}(a q)$ . (Do not do any numerical calculations.)

[ 2 ]

(b)

In the original $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution at equilibrium, is the concentration of $\mathrm{HCO}_{3}^{-}(a q)$ greater than, less than, or equal to the concentration of $\mathrm{CO}_{3}{ }^{2-}(a q)$ ? Justify your answer.

[ 1 ]

(c)

The student needs to make a $\mathrm{CO}_{3}{ }^{2-} / \mathrm{HCO}_{3}{ }^{-}$ buffer. Is the $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution suitable for making a buffer with a pH of 6 ? Explain why or why not.

[ 1 ]

[Maximum number: 2]

Answer the following questions about magnesium.

(a)

Calculate the pH of the solution in beaker 2.

[ 1 ]

For the correct calculated value:

Point 05

$\begin{aligned} & \mathrm{pOH}=-\log \left(2.80 \times 10^{-4}\right)=3.553 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-3.553=10.447 \end{aligned}$

(b)

In a separate experiment, the student adds $\mathrm{HNO}_{3}(a q)$ to decrease the pH of a saturated solution containing undissolved $\mathrm{Mg}(\mathrm{OH})_{2}(s)$ . Does the amount of undissolved $\mathrm{Mg}(\mathrm{OH})_{2}(s)$ increase, decrease, or remain the same as the $\mathrm{HNO}_{3}(a q)$ is added? Justify your answer.

[ 1 ]

For the correct answer and justification:

Point 10

Decrease. The $\mathrm{H}^{+}$ from $\mathrm{HNO}_{3}(a q)$ will react with $\mathrm{OH}^{-}$ , decreasing [ $\mathrm{OH}^{-}$ ] and causing $Q<K_{s p}$ . As a result, more $\operatorname{Mg}(\mathrm{OH})_{2}(s)$ will dissolve until equilibrium is reestablished, resulting in less $\mathrm{Mg}(\mathrm{OH})_{2}(s)$ .

[Maximum number: 2]

Answer the following questions about the glycolate ion, $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{3}{ }^{-}$ , which acts as a base in aqueous solution. A Lewis diagram for the ion is provided.

(a)

At $25^{\circ} \mathrm{C}$ , a 2.5 M solution of glycolate is found to have $\left[\mathrm{OH}^{-}\right]=1.3 \times 10^{-5} M$ .

[ 2 ]

(i)

Calculate the value of $K_{b}$ for the glycolate ion.

[ 1 ]

For the correct calculated value:

Point 02

$K_{b}=\frac{\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{3}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{3}^{-}\right]}=\frac{\left(1.3 \times 10^{-5}\right)\left(1.3 \times 10^{-5}\right)}{\left(2.5-1.3 \times 10^{-5}\right)} \approx \frac{\left(1.3 \times 10^{-5}\right)^{2}}{(2.5)}=6.8 \times 10^{-11}$

(ii)

Using your answer to part B (i), calculate the value of $K_{a}$ for glycolic acid at 25°C.

[ 1 ]

For the correct calculated value, consistent with part B (i):

Point 03

$K_{a}=\frac{K_{w}}{K_{b}}=\frac{1.0 \times 10^{-14}}{6.8 \times 10^{-11}}=1.5 \times 10^{-4}$

[Maximum number: 4]

Methanoic acid, HCOOH, ionizes according to the equation above.

(a)

Write the expression for the equilibrium constant, $K_{a}$ , for the reaction.

[ 1 ]

For the correct expression: 1 point

K_{a}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{HCOO}^{-}\right]}{[\mathrm{HCOOH}]}

(b)

Calculate the pH of a 0.25 M solution of HCOOH.

[ 2 ]

For the correct calculated concentration of $\mathrm{H}_{3} \mathrm{O}^{+}$ : 1 point

\mathrm{HCOOH}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HCOO}^{-}

\begin{array}{cccc} I & 0.25 & 0 & 0 \\ C & -x & +x & +x \\ \hline E & 0.25-x & x & x \end{array}

Let $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=x$ , then $1.8 \times 10^{-4}=\frac{x^{2}}{(0.25-x)}$
Assume $x \ll 0.25$ , then $1.8 \times 10^{-4}=\frac{x^{2}}{0.25} \Rightarrow x=0.0067 M$
For the correct calculated value of pH: 1 point

\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log (0.0067)=2.17

(c)

In aqueous solution, the compound $\mathrm{H}_{2} \mathrm{NNH}_{2}$ reacts according to the equation above. A 50.0 mL sample of $0.25 M \mathrm{H}_{2} \mathrm{NNH}_{2}(a q)$ is combined with a 50.0 mL sample of 0.25 M HCOOH(a q).

[ 1 ]

(i)

Is the resulting solution acidic, basic, or neutral? Justify your answer.

When a catalyst is added to a solution of HCOOH(a q), the reaction represented by the following equation occurs.

\mathrm{HCOOH}(a q) \rightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)

[ 1 ]

For the correct answer and a valid justification: 1 point Acidic. The $\mathrm{K}_{a}$ of $\mathrm{H}_{2} \mathrm{NNH}_{3}{ }^{+}$ is greater than the $\mathrm{K}_{b}$ of $\mathrm{HCOO}^{-}$ , so the production of $\mathrm{H}_{3} \mathrm{O}^{+}$ (aq) occurs to a greater extent than the production of OH (aq). Total for part (d) 2 points

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Covered syllabus points

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