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Chapter 17: Problem 90

Show that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to \(\mathrm{p} K_{a}\) for the acid.

Short Answer

Expert verified
At the halfway point of a titration of a weak acid with a strong base, half of the weak acid has been neutralized by the strong base. Therefore, the concentrations of the weak acid HX and its conjugate base X- are equal at this point: \([HX] = [X^-]\). Plugging the concentrations of HX and X- into the Ka expression, we get \(K_a = [H^+]\). Taking the negative logarithm of both sides, we find that \(\mathrm{pKa} = \mathrm{pH}\). This demonstrates that the pH at the halfway point of a titration of a weak acid with a strong base is equal to the pKa for the acid.

Step by step solution

01

Define weak acids, strong bases, pH, and Ka

A weak acid is an acid that doesn't completely ionize in solution, meaning that not all of its molecules dissociate into H+ ions. A strong base, on the other hand, is a base that completely ionizes in solution, releasing OH- ions. The pH of a solution is a measure of its acidity or alkalinity, defined as pH = -log10[H+] where [H+] is the concentration of H+ ions in the solution. The pKa of an acid is the negative log of its Ka value, where Ka is the acid dissociation constant.
02

Explain the titration process and the equivalence point

Titration is an analytical technique used to determine the concentration of a solute in a solution. In the case of titrating a weak acid with a strong base, the weak acid is the solute, and the strong base is added gradually to the solution. The process continues until the number of moles of base added equals the initial number of moles of the acid, which is called the equivalence point.
03

Write the dissociation equation of the weak acid

In order to analyze the halfway point of the titration, we need the dissociation equation of the weak acid, HX, which can be written as: \[HX \rightleftharpoons H^+ + X^-\]
04

Determine the dissociation constants

The dissociation constant for the weak acid can be expressed as Ka: \[K_a = \frac{[H^+][X^-]}{[HX]}\] and the pH is related to the concentration of hydrogen ions [H+]: \[\mathrm{pH} = -\log_{10} [H^+]\]
05

Analyze the halfway point of the titration

At the halfway point of the titration, half of the weak acid has been neutralized by the strong base. Therefore, the concentrations of HX and its conjugate base X- are equal at this point: \[[HX] = [X^-]\]
06

Show that pH is equal to pKa at the halfway point

At the halfway point, we can plug the concentrations of HX and X- into the Ka expression: \[K_a = \frac{[H^+][X^-]}{[HX]}\] Since [HX] = [X-]: \[K_a = [H^+]\] Now, we can take the negative logarithm of both sides: \[-\log_{10}{K_a} = -\log_{10}{[H^+]}\] This gives us the relation between the pKa and the pH at the halfway point: \[\mathrm{pKa} = \mathrm{pH}\] This shows that the pH at the halfway point of a titration of a weak acid with a strong base is equal to the pKa for the acid, as required.

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Most popular questions from this chapter

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

(a) Calculate the percent ionization of \(0.125 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right)\). (b) Calculate the percent ionization of \(0.125 \mathrm{M}\) lactic acid in a solution containing \(0.0075 \mathrm{M}\) sodium lactate.

A 20.0-mL sample of \(0.200 \mathrm{M}\) HBr solution is titrated with \(0.200 \mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) of the solution after the following volumes of base have been added: (a) \(15.0 \mathrm{~mL}\) (b) \(19.9 \mathrm{~mL}\) (c) \(20.0 \mathrm{~mL},\) (d) \(20.1 \mathrm{~mL},\) (e) \(35.0 \mathrm{~mL}\)

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C} ?\)

You have to prepare a pH 5.00 buffer, and you have the following \(0.10 \mathrm{M}\) solutions available: \(\mathrm{HCOOH}, \mathrm{HCOONa}, \mathrm{CH}_{3} \mathrm{COOH},\) \(\mathrm{CH}_{3} \mathrm{COONa}, \mathrm{HCN},\) and \(\mathrm{NaCN}\). Which solutions would you use? How many milliliters of each solution would you use to make approximately a liter of the buffer?
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