Home / AP Chemistry: 8.2 pH and pOH of Strong Acids and Bases – Exam Style questions with Answer- FRQ

AP Chemistry: 8.2 pH and pOH of Strong Acids and Bases – Exam Style questions with Answer- FRQ

Question

                                             \(CH_3CH_2COOH(aq) + H_2O(l) \rightleftharpoons CH_3CH_2COO^ ̄(aq) + H2O^{+}(aq) \)

Propanoic acid,   \( CH_3CH_2COOH\),   is a carboxylic acid that reacts with water according to the equation above. At 25°C the pH of a 50.0 mL sample of 0.20 M   \(CH_3CH_2COOH \)   is 2.79.

(a) Identify a Brønsted-Lowry conjugate acid-base pair in the reaction. Clearly label which is the acid and which is the base.

(b) Determine the value of K for propanoic acid at 25°C.

(c) For each of the following statements, determine whether the statement is true or false. In each case, explain the reasoning that supports your answer. (i) The pH of a solution prepared by mixing the 50.0 mL sample of 0.20 M \(CH_3CH_2COOH\) with a 50.0 mL sample of 0.20 M NaOH is 7.00. (ii) If the pH of a hydrochloric acid solution is the same as the pH of a propanoic acid solution, then the molar concentration of the hydrochloric acid solution must be less than the molar concentration of the propanoic acid solution. A student is given the task of determining the concentration of a propanoic acid solution of unknown concentration. A 0.173 M NaOH solution is available to use as the titrant. The student uses a 25.00 mL volumetric pipet to deliver the propanoic acid solution to a clean, dry flask. After adding an appropriate indicator to the flask, the student titrates the solution with the 0.173 M NaOH, reaching the end point after 20.52 mL of the base solution has been added.

(d) Calculate the molarity of the propanoic acid solution.

(e) The student is asked to redesign the experiment to determine the concentration of a butanoic acid solution instead of a propanoic acid solution. For butanoic acid the value of pK is 4.83. The student claims that a different indicator will be required to determine the equivalence point of the titration accurately. Based on your response to part (b), do you agree with the student’s claim? Justify your answer.

▶️Answer/Explanation

(a)  \(CH_3CH_2COOH\) and \(CH _3CH_2COO ^-\)

               acid                                               base

                                           OR

                     \(H_2O^+\)  and   \(H_2O\)

                                   acid                  base

(b) \([H_3O^+]=10^{-pH}=10^{-2.79}=1.6\times 10^{-3}\) M

         \([CH _3CH_2COO^ ̄]=[H_3O^+]\)

AND

\([CH _3CH_2COOH]=20M-[H_3O^+]\) , OR \([CH _3CH_2COOH]\approx 0.20 M\)

\(K_a=\frac{[CH _3CH_2COO^ ̄][H_3O^+]}{[CH _3CH_2COOH]}=\frac{(1.6\times 10^{-3}M)^2}{0.20 M}\)

c.(i) False. The conjugate base of a weak acid undergoes hydrolysis (see equation below) at equivalence to form a solution with a pH>7.

                \(CH _3CH_2COO^ ̄ + H_2O\rightleftharpoons CH_2CH_2COOH + OH ^-\)

c.(ii) True. HCI is a strong acid that ionizes completely. Fewer moles of HCI are needed to produce the same [H3O+] as the propanoic acid solution, which only partially ionizes.

(d) Let x = moles of propanoic acid

then \(x =  (0.02052 L NaOH) \times  \frac{0.173 mol NaOH}{1 L NaOH}\times \frac{1 mol acid }{1 mol NaOH} = 3.55 \times  10^{-3} mol propanoic acid \Rightarrow \frac{3.55 \times  10^{-3} mol acid}{ 0.02500 L acid}  =0.142 M\)

OR

Since \(CH_3CH_2COOH\) is monoprotic and, at the equivalence point, moles \(H^+ = moles OH^-\), then

\(M_AV_A=M_BV_B\)

\(M_A=\frac{M_BV_B}{ V_A}= \frac{(0.173 M NaOH)(20.52 mL NaOH)}{ 25.00 mL acid} =0.142 M\)

(e) Disagree with the student’s claim From part (b) above, \(pK_a\), for propanoic acid is \(log(1.3 × 10^{_5}) = 4.89\). Because 4.83 is so close to 4.89, the pH at the equivalence point in the titration of butanoic acid should be close enough to the pH in the titration of propanoic acid to make the original indicator appropriate for the titration of butanoic acid.

Question.

                                             \(H_{3}BO_{3}(aq)+4HF(g)\rightarrow H_{3}O^{+}(aq)+BF_{4}^{-}(aq)+2H_{2}O(l)\)

Tetrafluoroboric acid is a strong acid with the formula\( HBF_4\). The acid can be prepared by reacting the weak acid \(H_3BO_3\) (molar mass 61.83 g/mol) with HF according to the equation above.

(a) To prepare a solution of \(BF_4^{-}\)(aq), HF(g) is bubbled into a solution containing 50.0 g of\( H_3BO_3\) in a 1 L reaction vessel.
(i) Calculate the maximum number of moles of\( BF_4^{-}\)(aq) that can be produced.

(ii) Calculate the number of liters of HF(g), measured at 273 K and 1.00 atm, that will be consumed if all the\( H_3BO_3\) reacts.
(iii) Will the pH of the solution increase, decrease, or remain the same during the course of the reaction? Justify your answer. In another experiment, a 0.150 M \(BF_4^-\)(aq) solution is prepared by dissolving \(NaBF_4\)(s) in distilled water. The\( BF_{4}^-\)(aq) ions in the solution slowly react with H2O(l) in the reversible reaction represented below.

                                                \(BF_4^-(aq)+H_{2}O(l)\rightleftharpoons BF_{3}OH^{-}(aq)+HF(aq)\)

The concentration of HF is monitored over time, as shown in the graph below.

[HF] reaches a constant value of 0.0174 M when the reaction reaches equilibrium. For the forward reaction, the rate law is rate = kf \([BF_4^{-}\)]. The value of the rate constant kf was experimentally determined to be \(9.00\times 10^{-4}min^{-1}\).

(b) Calculate the rate of the forward reaction after 600. minutes. Include units with your answer. The rate law for the reverse reaction is rate = \(kr [BF_3OH^-]\)[HF].
(c) A student claims that the initial rate of the reverse reaction is equal to zero. Do you agree or disagree with this claim? Justify your answer in terms of the rate law for the reverse reaction.
(d) At equilibrium the forward and reverse reaction rates are equal. Calculate the value of the rate constant for the reverse reaction.

                                                                   \(H_{3}BO_{3}(aq)+4HF(g)\rightarrow H_{3}O^{+}(aq)+BF_{4}^{-}(aq)+2H_{2}O(l)\)

Tetrafluoroboric acid is a strong acid with the formula \(HBF_4\). The acid can be prepared by reacting the weak acid \(H_3BO_3\) (molar mass 61.83 g/mol) with HF according to the equation above.

▶️Answer/Explanation

a(i) \(50.0gH_{3}BO_{3}\times \frac{1molH_{3}BO_{3}}{61.83g}\times \frac{1molBF_{4}^{-}}{1molH_{3}BO_{3}}\)

a(ii)\( 0.809 mol \times 4 = 3.24 mol\)

           PV= nRT

          \(V=\frac{nRT}{p}=\frac{(3.24mol)(0.08206Latm mol^{-1}K^{-1})(273K)}{1.00atm}\)
           =72.6L

                  OR
        \(0.809 mol \times 4 = 3.24 mol\)
      \(\frac{22.4L}{1mol}\times 3.24mol=72.6 L\)

a(iii) As the reaction proceeds,\( H_3O^{+}\) is produced, so the pH
will decrease.

(b)  \([BF_4^{-}] = 0.150 M – 0.0174 M = 0.133 M\)
\(rate = (9.00\times 10^{-4}min^{-1})(0.133M)=1.20\times 10^{-4}M min^{-1}\)

(c) Agree. The initial concentration of each product is zero, so the initial rate of the reverse reaction is zero.

(d ) 

Question

 A student prepares three solutions, X, Y, and Z, as described in the table above. The values of \(K_a\) for the acidic species in the solutions are given in the table below.

(a) Using the information above, write the letters of the solutions in the boxes below to rank the solutions in order of increasing pH. Explain your reasoning for the ranking.

(b) Does the pH of solution Y increase, decrease, or remain the same when 100 mL of water is added? Justify your answer.
(c) The student adds 0.0010 mol of NaOH(s) to solution Y, and adds 0.0010 mol of NaOH(s) to solution Z. Assume that the volume of each solution does not change when the NaOH(s) is added. The pH of solution Y changes much more than the pH of solution Z changes. Explain this observation.

▶️Answer/Explanation

(a) Solution Y is a strong acid solution with a very low pH. Solution Z is a buffer solution with pKa = 4.74 = pH. Solution X is a neutral solution created from equimolar amounts of a strong acid and a strong base that react in a 1:1 ratio.

(b) The pH of the solution increases. The addition of water will decrease\( [H^+]\); therefore, the pH will increase.

(c) Solution Z is a buffer system (composed of a weak acid and its conjugate base), whereas solution Y is not a buffer.

Scroll to Top