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Chapter 17: Problem 44

A 20.0-mL sample of \(0.150 \mathrm{MKOH}\) is titrated with \(0.125 \mathrm{M}\) \(\mathrm{HClO}_{4}\) solution. Calculate the pH after the following volumes of acid have been added: $(\mathbf{a}) 20.0 \mathrm{~mL},(\mathbf{b}) 23.0 \mathrm{~mL},\( (c) \)24.0 \mathrm{~mL}\( (d) \)25.0 \mathrm{~mL},\( (e) \)30.0 \mathrm{~mL}$.

Short Answer

Expert verified
The pH at each point of titration is: (a) 13.09, (b) 12.52, (c) 7.00, (d) 4.48, and (e) 3.93.

Step by step solution

01

Write the reaction equation between the base and the acid

The reaction between the strong base KOH and the strong acid HClO₄ can be represented by the following balanced chemical equation: \(KOH + HClO_4 \rightarrow KClO_4 + H_2O\)
02

Calculate the initial moles of KOH

We are given a 20.0 mL sample of 0.150 M KOH. To find the initial moles of KOH, use the formula: moles = Molarity × Volume in liters moles of KOH = 0.150 M × 0.020 L = 0.003 mol
03

Calculate the moles of HClO₄ added for each volume

The given concentration of HClO₄ is 0.125 M. We will calculate the moles of HClO₄ for the respective added volumes. (a) 20.0 mL: moles of HClO₄ = 0.125 M × 0.020 L = 0.00250 mol (b) 23.0 mL: moles of HClO₄ = 0.125 M × 0.023 L = 0.002875 mol (c) 24.0 mL: moles of HClO₄ = 0.125 M × 0.024 L = 0.00300 mol (d) 25.0 mL: moles of HClO₄ = 0.125 M × 0.025 L = 0.003125 mol (e) 30.0 mL: moles of HClO₄ = 0.125 M × 0.030 L = 0.00375 mol
04

Determine the pH at each point of titration

We know the relation between moles of KOH and moles of HClO₄, pH can be calculated as follows: (a) 0.003 mol KOH - 0.00250 mol HClO₄ = 0.0005 mol KOH left pH = 14 - pOH pOH = -log(OH⁻ molarity) pH = 14 - pOH(KOH) = 14 - (-log((0.0005 mol)/(0.040 L))) = 13.09 (b) 0.003 mol KOH - 0.002875 mol HClO₄ = 0.000125 mol KOH left pH = 14 - pOH(KOH) = 14 - (-log((0.000125 mol)/(0.043 L))) = 12.52 (c) 0.003000 mol KOH - 0.003000 mol HClO₄ = 0 (neutralization point) pH = 7 (d) 0.003000 mol KOH - 0.003125 mol HClO₄ = 0.000125 mol HClO₄ left (excess acid) pH = -log(H⁺ molarity) pH = -log((0.000125 mol)/(0.045 L)) = 4.48 (e) 0.003000 mol KOH - 0.003750 mol HClO₄ = 0.00075 mol HClO₄ left (excess acid) pH = -log((0.00075 mol)/(0.050 L)) = 3.93 The pH at each point of titration is: (a) 13.09 (b) 12.52 (c) 7.00 (d) 4.48 (e) 3.93

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Most popular questions from this chapter

A buffer is prepared by adding \(15.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to \(500 \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) solution. (a) Determine the pH of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer.

The solubility of \(\mathrm{CaCO}_{3}\) is pH dependent. (a) Calculate the molar solubility of \(\mathrm{CaCO}_{3}\left(K_{s p}=4.5 \times 10^{-9}\right)\) neglecting the acid-base character of the carbonate ion. (b) Use the \(K_{b}\) expression for the \(\mathrm{CO}_{3}^{2-}\) ion to determine the equilibrium constant for the reaction $$ \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons $$ (c) If we assume that the only sources of $\mathrm{Ca}^{2+}, \mathrm{HCO}_{3}^{-},\( and \)\mathrm{OH}^{-}$ ions are from the dissolution of \(\mathrm{CaCO}_{3},\) what is the molar solubility of \(\mathrm{CaCO}_{3}\) using the equilibrium expression from part (b)? (d) What is the molar solubility of \(\mathrm{CaCO}_{3}\) at the \(\mathrm{pH}\) of the ocean (8.3)\(?(\mathbf{e})\) If the \(\mathrm{pH}\) is buffered at \(7.5,\) what is the molar solubility of \(\mathrm{CaCO}_{3} ?\)

A solution contains \(1.0 \times 10^{-4} \mathrm{Ca}^{2+}(a q)\) and $1.0 \times 10^{-4}\( \)\mathrm{La}^{3+}(a q) .\( If \)\mathrm{NaF}$ is added, will \(\mathrm{CaF}_{2}\left(K_{s p}=3.9 \times 10^{-11}\right)\) or \(\mathrm{LaF}_{3}\left(K_{s p}=2 \times 10^{-19}\right)\) precipitate first? Specify the concentration of \(\mathrm{F}^{-}(a q)\) needed to begin precipitation.

A 10.0-mL sample of \(0.250 \mathrm{MHNO}_{3}\) solution is titrated with $0.100 M$ KOH solution. Calculate the pH of the solution after the following volumes of base have been added: (a) \(20.0 \mathrm{~mL}\), (b) \(24.9 \mathrm{~mL}\) (c) \(25.0 \mathrm{~mL}\) (d) \(25.1 \mathrm{~mL}\), (e) \(30.0 \mathrm{~mL}\).

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) is adjusted to \(8.0 ?(\mathbf{b})\) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when \(100 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of $5.0 \times 10^{-2} \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$ solution?
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