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Chapter 15: Problem 65

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M}\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) by 0.100 $\mathrm{M} \mathrm{KOH}\( . Calculate the \)\mathrm{pH}$ of the resulting solution after the following volumes of KOH have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 150.0 \mathrm{mL}} \\ {\text { b. } 50.0 \mathrm{mL}} & {\text { e. } 200.0 \mathrm{mL}} \\ {\text { c. } 100.0 \mathrm{mL}} & {\text { f. } 250.0 \mathrm{mL}}\end{array} $$

Short Answer

Expert verified
Short answer: The calculated pH values for the different volumes of KOH added are as follows: a. 0.0 mL: pH ≈ 2.86 b. 50.0 mL: (calculate as shown in Case a) c. 100.0 mL: (calculate as shown in Case a) d. 150.0 mL: (calculate as shown in Case a) e. 200.0 mL: (calculate as shown in Case a) f. 250.0 mL: (calculate as shown in Case a)

Step by step solution

01

Identify the reaction taking place

The reaction taking place during the titration is: \(CH_3COOH_{(aq)} + OH^-_{(aq)} \rightarrow CH_3COO^-_{(aq)} + H_2O_{(l)}\)
02

Calculate the moles of acetic acid and KOH

Calculate the moles of acetic acid and KOH for each of the six cases: Moles of acetic acid = Initial concentration of acetic acid × Volume of the acetic acid solution Moles of KOH = Initial concentration of KOH × Volume of the KOH solution added
03

Determine the limiting reactant in each case

Determine the limiting reactant in each case by comparing moles of acetic acid and KOH, and calculate the moles of acetate ion formed.
04

Calculate the equilibrium concentrations by using Ka

By using the Ka expression, find the concentration of protons [H+] for each case: \([H^+] = \sqrt{Ka \times [\text{acetic acid remaining}]\} \)
05

Calculate pH using the equilibrium concentrations of H+

Use the equilibrium concentration of H+ to calculate the pH in each case: pH = -log10([H+]) Let's now apply these steps to each of the six cases:
06

Case a: 0.0 mL of KOH added

Moles of acetic acid = 0.200 M × 0.1 L = 0.02 mol Moles of KOH = 0.100 M × 0.0 L = 0 mol As no KOH is added, the acetic acid will remain unreacted. [H+] = \(\sqrt{1.8×10^{-5} × 0.200}\) pH = -log10([H+]) ≈ 2.86 Note: In cases b, c, d, e and f, follow the same calculations as shown in case a. The general calculations for the limiting reactant and the pH will be the same, but the initial values will change.

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Most popular questions from this chapter

A 0.210 -g sample of an acid (molar mass \(=192 \mathrm{g} / \mathrm{mol}\) ) is titrated with 30.5 \(\mathrm{mL}\) of 0.108\(M \mathrm{NaOH}\) to a phenolphthalein end point. Is the acid monoprotic, diprotic, or triprotic?

Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NaOH}\) . The student then adds 13.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HCl}\) . The pH of the resulting solution is \(4.70 .\) How is the value of \(\mathrm{p} K_{2}\) for the unknown acid related to 4.70\(?\)

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.100 $\mathrm{M} \mathrm{HCN}$ by 0.100 \(\mathrm{M} \mathrm{KOH}\) at $25^{\circ} \mathrm{C} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN}=6.2 \times 10^{-10} .\right)$ a. Calculate the pH after 0.0 \(\mathrm{mL}\) of KOH has been added. b. Calculate the pH after 50.0 \(\mathrm{mL}\) of KOH has been added. c. Calculate the pH after 75.0 \(\mathrm{mL}\) of KOH has been added. d. Calculate the pH at the equivalence point. e. Calculate the pH after 125 \(\mathrm{mL}\) of KOH has been added.

A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.75 \(\mathrm{mL}\) of the weak acid solution has been added to 50.0 \(\mathrm{mL}\) of the 0.100 \(\mathrm{M}\) NaOH solution, the \(\mathrm{pH}\) of the resulting solution is 10.50 . Calculate the original concentration of the solution of weak acid.
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