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Chapter 17: Problem 92

Mathematically prove 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 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 reacted, and we can represent the concentrations as equal amounts of the undissociated weak acid (A) and its conjugate base (A-), both with concentration x. The equilibrium constant (Ka) is given by \[ K_a = \frac{[H^+][A^-]}{[A]} \]. Since [H+] = x as well, we have \[ K_a = x \]. Using the relationships \( pK_a = - \log_{10}K_a \) and \( pH = - \log_{10}[H^+] \), we find that \( pK_a = pH \) at the halfway point of the titration.

Step by step solution

01

Write the dissociation equation for the weak acid

A weak acid (A) in water will dissociate into a hydrogen ion (H+) and the conjugate base (A-) in the following equilibrium reaction: \[ A(aq) \rightleftharpoons H^+(aq) + A^-(aq) \]
02

Write the expression for the acid dissociation constant (Ka)

The equilibrium constant (Ka) for the weak acid dissociation is given by the concentrations of the products (H+ and A-) divided by the concentration of the weak acid (A): \[ K_a = \frac{[H^+][A^-]}{[A]} \]
03

Determine the concentration of species at the halfway point

At the halfway point of the titration, half of the weak acid (A) has reacted to form an equal amount of its conjugate base (A-). Therefore, we can express the concentrations of A and A- in terms of x, where x is half of the initial concentration of the weak acid (A): \[ [A] = x \] \[ [A^-] = x \] Also, the hydrogen ion concentration ([H+]) will be equal to the concentration of the conjugate base (A-), which we established as x: \[ [H^+] = x \]
04

Substitute the concentrations into the Ka expression

Now, we will substitute the concentrations of A, A-, and H+ from step 3 into the Ka expression from step 2: \[ K_a = \frac{[H^+][A^-]}{[A]} = \frac{x \cdot x}{x} = x \]
05

Relating pKa to the pH at the halfway point

Now, we will calculate the pKa and pH, using the relationships: \[ pK_a = - \log_{10}K_a \] \[ pH = - \log_{10}[H^+] \] Since Ka = x and [H+] = x, we can say: \[ pK_a = - \log_{10}(x) \] \[ pH = - \log_{10}(x) \] Thus, at the halfway point of a titration of a weak acid with a strong base, the pH is equal to the pKa of the acid: \[ pH = pK_a \]

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Most popular questions from this chapter

You have to prepare a \(\mathrm{pH}=5.00\) buffer, and you have the following 0.10 \(\mathrm{M}\) solutions available: HCOOH, 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 1 \(\mathrm{L}\) of the buffer?

Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) ZnCO \(_{3}\) \(\mathbf{b} ) \mathrm{ZnS},(\mathbf{c}) \mathrm{BiI}_{3},(\mathbf{d}) \mathrm{AgCN},(\mathbf{e}) \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2} ?\)

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 \(\mathrm{mL}\) \((\mathbf{b}) 19.9 \mathrm{mL},(\mathbf{c}) 20.0 \mathrm{mL},(\mathbf{d}) 20.1 \mathrm{mL},(\mathbf{e}) 35.0 \mathrm{mL}\)

(a) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood of pH 7.4 ? (b) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in an exhausted marathon runner whose blood pH is 7.1\(?\)

Lead(II) carbonate, PbCO \(_{3},\) is one of the components of the passivating layer that forms inside lead pipes.(a) If the \(K_{s p}\) for \(\mathrm{PbCO}_{3}\) is \(7.4 \times 10^{-14}\) what is the molarity of \(\mathrm{Pb}^{2+}\) in a saturated solution of lead(II) carbonate? (b) What is the concentration in ppb of \(\mathrm{Pb}^{2+}\) ions in a saturated solution? (c) Will the solubility of \(\mathrm{PbCO}_{3}\) increase or decrease as the \(\mathrm{pH}\) is lowered? \((\boldsymbol{d} )\)The EPA threshold for acceptable levels of lead ions in water is 15 ppb. Does a saturated solution of lead(II) carbonate produce a solution that exceeds the EPA limit?
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