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Chapter 15: Problem 63

Consider the titration of 40.0 \(\mathrm{mL}\) of 0.200 $\mathrm{M} \mathrm{HClO}_{4}$ by 0.100 \(\mathrm{M}\) KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 80.0 \mathrm{mL}} \\ {\text { b. } 10.0 \mathrm{mL}} & {\text { e. } 100.0 \mathrm{mL}} \\ {\text { c. } 40.0 \mathrm{mL}}\end{array} $$

Short Answer

Expert verified
The pH of the titration solution after different volumes of KOH have been added is as follows: a. 0.0 mL of KOH added: pH = 0.699 b. 10.0 mL of KOH added: pH = 0.853 c. 40.0 mL of KOH added: pH = 1.301 d. 80.0 mL of KOH added: pH = 7 e. 100.0 mL of KOH added: pH = 12.155

Step by step solution

01

Find the initial moles of HClO₄ and KOH

Since we need to determine the concentrations of HClO₄ and KOH throughout the titration, it is helpful to first find the initial amount of moles of these two chemicals. - Moles of HClO₄ = Molarity × Volume = (0.200 mol/L) × (40.0 mL)×(1 L/1000 mL) = 0.00800 mol - Moles of KOH (initially) = 0 mol
02

Determine the moles of HClO₄ and KOH at each volume

For each volume of KOH added (a = 0.0 mL; b = 10.0 mL; c = 40.0 mL; d = 80.0 mL; e = 100.0 mL), we will determine the moles of HClO₄ and KOH at that point. a. 0.0 mL of KOH added: - Moles of HClO₄ = 0.00800 mol - Moles of KOH = 0 mol b. 10.0 mL of KOH added: - Moles of KOH added = (0.100 mol/L) × (10.0 mL) × (1 L/1000 mL) = 0.00100 mol - Moles of HClO₄ remaining = 0.00800 mol - 0.00100 mol = 0.00700 mol - Moles of KOH remaining = 0 mol (all react with HClO₄) c. 40.0 mL of KOH added: - Moles of KOH added = (0.100 mol/L) × (40.0 mL) × (1 L/1000 mL) = 0.00400 mol - Moles of HClO₄ remaining = 0.00800 mol - 0.00400 mol = 0.00400 mol - Moles of KOH remaining = 0 mol (all react with HClO₄) d. 80.0 mL of KOH added: - Moles of KOH added = (0.100 mol/L) × (80.0 mL) × (1 L/1000 mL) = 0.00800 mol - Moles of HClO₄ remaining = 0 mol (moire of KOH react) - Moles of KOH remaining = 0.00800 mol - 0.00800 mol = 0 mol e. 100.0 mL of KOH added: - Moles of KOH added = (0.100 mol/L) × (100.0 mL) × (1 L/1000 mL) = 0.01000 mol - Moles of HClO₄ remaining = 0 mol (moire of KOH react) - Moles of KOH remaining = 0.01000 mol - 0.00800 mol = 0.00200 mol
03

Calculate the pH for each volume

For each volume, we will calculate the pH using the corresponding moles of HClO₄ and KOH. a. 0.0 mL of KOH added: - [H₃O⁺] = moles of HClO₄ / total volume = 0.00800 mol / 0.04000 L = 0.200 M - pH = -log(0.200) = 0.699 b. 10.0 mL of KOH added: - [H₃O⁺] = moles of HClO₄ / total volume = 0.00700 mol / 0.05000 L = 0.140 M - pH = -log(0.140) = 0.853 c. 40.0 mL of KOH added: - [H₃O⁺] = moles of HClO₄ / total volume = 0.00400 mol / 0.08000 L = 0.05000 M - pH = -log(0.05000) = 1.3010 d. 80.0 mL of KOH added (at equivalence point: [H₃O⁺] = [OH⁻]): - pH = 7 e. 100.0 mL of KOH added: - [OH⁻] = moles of KOH remaining / total volume = 0.00200 mol / 0.14000 L = 0.01429 M - pOH = -log(0.01429) = 1.845 - pH = 14 - pOH = 14 - 1.845 = 12.155
04

Summary

We have calculated the pH of the titration solution after different volumes of KOH have been added: a. 0.0 mL of KOH added: pH = 0.699 b. 10.0 mL of KOH added: pH = 0.853 c. 40.0 mL of KOH added: pH = 1.3010 d. 80.0 mL of KOH added: pH = 7 e. 100.0 mL of KOH added: pH = 12.155

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Most popular questions from this chapter

Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100$M \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\( with 0.100 \)\mathrm{M}$ \(\mathrm{HCl} .\)

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

When a diprotic acid, \(\mathrm{H}_{2} \mathrm{A},\) is titrated with NaOH, the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at 100.0 mL NaOH added, what volume of NaOH added corresponds to the second equivalence point? b. For the following volumes of NaOH added, list the major species present after the OH- reacts completely. $$ \begin{array}{l}{\text { i. } 0 \mathrm{mL} \text { NaOH added }} \\ {\text { i. between } 0 \text { and } 100.0 \mathrm{mL} \text { NaOH added }}\end{array} $$ $$ \begin{array}{l}{\text { iii. } 100.0 \text { mL NaOH added }} \\ {\text { iv. between } 100.0 \text { and } 200.0 \mathrm{mL} \text { NaOH added }} \\\ {\text { v. } 200.0 \mathrm{mL} \text { NaOH added }}\end{array} $$ $$ 200.0 \mathrm{mL} $$ c. If the pH at 50.0 \(\mathrm{mL}\) NaOH added is 4.0 and the pH at 150.0 \(\mathrm{mL}\) . NaOH added is \(8.0,\) determine the values \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for the diprotic acid.

Th pH of blood is steady at a value of approximately 7.4 as a result of the following equilibrium reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \leftrightharpoons \mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q) $$ The actual buffer system in blood is made up of $\mathrm{H}_{2} \mathrm{CO}_{3}\( and \)\mathrm{HCO}_{3}$ - One way the body keeps the pH of blood at 7.4 is by regulating breathing. Under what blood ph conditions will the body increase breathing and under what blood pH conditions will the body decrease breathing? Explain.

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intra- cellular fluids at pH values generally between 7.1 and \(7.2 .\) a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) inintracellular fluid at \(\mathrm{pH}=7.15 ?\) $$ \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8} $$ b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the pH of intracellular fluid? $$ \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{2}=7.5 \times 10^{-3} $$
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