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Chapter 14: Problem 99

A sample of a certain monoprotic weak acid was dissolved in water and titrated with 0.125 \(M\) NaOH, requiring \(16.00 \mathrm{mL}\) to reach the equivalence point. During the titration, the pH after adding \(2.00 \mathrm{mL}\) NaOH was \(6.912 .\) Calculate \(K_{\mathrm{a}}\) for the weak acid.

Short Answer

Expert verified
The Ka value for the weak acid is approximately \(1.74 \times 10^{-9}\).

Step by step solution

01

Determine the moles of OH- in the added NaOH

To calculate the moles of OH- in the added NaOH, we can use the given concentration and volume: Moles of OH- = \(concentration \times volume\) Moles of OH- = \(0.125 \frac{mol}{L} \times 0.002 L\) Moles of OH- = \(0.00025 mol\)
02

Calculate the moles of the weak acid present in the sample

Since we know that it required 16.00 mL of 0.125 M NaOH to reach the equivalence point, we can determine the moles of the weak acid in the sample: Moles of weak acid = moles of OH- (at equivalence point) Moles of weak acid = \(0.125 \frac{mol}{L} \times 0.016 L\) Moles of weak acid = \(0.002 mol\)
03

Calculate the initial concentrations of the weak acid and its conjugate base

After adding 2.00 mL of 0.125 M NaOH, the moles of the weak acid and its conjugate base will be: Moles of weak acid remaining = \(0.002 mol - 0.00025 mol = 0.00175 mol\) Moles of conjugate base formed = \(0.00025 mol\) Now, we can calculate the concentrations of the weak acid and its conjugate base: Initial concentration of the weak acid = \(\frac{0.00175 mol}{0.018 L} = 0.09722\frac{mol}{L}\) Initial concentration of the conjugate base = \(\frac{0.00025 mol}{0.018 L} = 0.01389\frac{mol}{L}\)
04

Use the provided pH and an equilibrium expression to determine the Ka value

From the given pH, we can determine the hydrogen ion concentration: pH = 6.912 \[H^+ = 10^{-pH} = 10^{-6.912} = 1.209 \times 10^{-7}\] Now, we can use the Ka expression to calculate Ka value: For a weak acid, HA, and its conjugate base, A-, the Ka expression is: \[K_a = \frac{[H^+][A^-]}{[HA]}\] Here, H+ is the hydrogen ion concentration, A- is the conjugate base concentration, and HA is the weak acid concentration. We can plug in the values: \[K_a = \frac{(1.209 \times 10^{-7})(0.01389)}{0.09722}\] \[K_a = 1.74 \times 10^{-9}\] So, the Ka value for the weak acid is approximately \(1.74 \times 10^{-9}\).

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Most popular questions from this chapter

Sketch two pH curves, one for the titration of a weak acid with a strong base and one for a strong acid with a strong base. How are they similar? How are they different? Account for the similarities and the differences.

Calculate the pH of each of the following solutions. a. \(0.100 M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) b. \(0.100 M\) sodium propanoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 M \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 M\) \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)

Calculate the volume of \(1.50 \times 10^{-2} \mathrm{M}\) NaOH that must be added to \(500.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) HCl to give a solution that has \(\mathrm{pH}=2.15\).

Sketch a pH curve for the titration of a weak acid (HA) with a strong base (NaOH). List the major species, and explain how you would go about calculating the pH of the solution at various points, including the halfway point and the equivalence point.

Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of each solution added change the effectiveness of the buffer?
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