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Chapter 17: Problem 83

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 \(\mathrm{pK}_{a}\) for the acid.

Short Answer

Expert verified
At the halfway point of a titration of a weak acid with a strong base, half the moles of weak acid have reacted with the strong base, forming an equal number of moles of its conjugate base. By using the acid dissociation constant (Ka) definition, we find that the concentration of \(H^{+}\) ions is equal to the Ka of the acid. This allows us to derive that the pH of the solution at the halfway point is equal to the pKa of the weak acid: \[pH = -\log{[H^{+}]} = -\log{K_{a}} = pK_{a}\].

Step by step solution

01

Write down the equations for the weak acid and strong base reactions

To understand the reactions occurring in the solution, let's identify the weak acid (HA) and the strong base (OH-), and write their respective equilibrium reactions: Weak acid dissociation: \[HA \rightleftharpoons H^{+} + A^{-}\] Strong base dissociation: \[OH^{-} \rightarrow O^{2-} + H^{+}\] In both reactions, we are interested in the production of \(H^{+}\) ions because that determines the pH of the solution.
02

Write down the definition of the acid dissociation constant

The acid dissociation constant (Ka) is defined as the equilibrium constant for the reaction of the weak acid (HA) with water: \[K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\] We will use this definition to find the concentration of \(H^{+}\) ions in the solution at the halfway point of the titration.
03

Determine the concentration of the weak acid at the halfway point

At the halfway point of the titration, the moles of strong base added equal half the moles of weak acid initially present in the solution. This means that half the moles of weak acid have reacted with the strong base, forming an equal number of moles of its conjugate base (A-). Thus, at the halfway point: \[[HA] = [A^{-}]\] \[x = \frac{[HA_{o}]}{2}\] Where \(x\) represents the moles of the weak acid reacted, and \([HA_{o}]\) is the initial concentration of the weak acid.
04

Use the Ka definition to find the concentration of H+ ions

Since we know that at the halfway point, \([HA] = [A^{-}]\), we can use the acid dissociation constant equation to find the \(H^{+}\) ion concentration: \[K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\] \[K_{a} = \frac{[H^{+}](\frac{[HA_{o}]}{2})}{\frac{[HA_{o}]}{2}}\] Simplifying the equation: \[[H^{+}] = K_{a}\]
05

Calculate the pH and pKa

Now that we have the concentration of \(H^{+}\) ions, we can use the definition of pH to find the pH of the solution: \[pH = -\log{[H^{+]}\] Similarly, we can define pKa as: \[pK_{a} = -\log{K_{a}}\] Since we know that \([H^{+}] = K_{a}\), the pH of the solution at the halfway point of the titration should be equal to the pKa of the weak acid: \[pH = -\log{[H^{+}]} = -\log{K_{a}} = pK_{a}\] Thus, we have shown 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.

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Most popular questions from this chapter

(a) Write the net ionic equation for the reaction that occurs when a solution of hydrochloric acid \((\mathrm{HCl})\) is mixed with a solution of sodium formate \(\left(\mathrm{NaCHO}_{2}\right)\). (b) Calculate the equilibrium constant for this reaction. (c) Calculate the equilibrium concentrations of \(\mathrm{Na}^{+}, \mathrm{Cl}^{-}\) \(\mathrm{H}^{+}, \mathrm{CHO}_{2}^{-}\), and \(\mathrm{HCHO}_{2}\) when \(50.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{HCl}\) is mixed with \(50.0 \mathrm{~m}\) L of \(0.15 \mathrm{M} \mathrm{NaCHO}_{2}\).

Calculate the \(\mathrm{pH}\) at the equivalence point for titrating \(0.200 \mathrm{M}\) solutions of each of the following bases with \(0.200 \mathrm{M} \mathrm{HBr}\) : (a) sodium hydroxide \((\mathrm{NaOH})\), (b) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right)\), (c) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\).

Suppose that a 10-mL sample of a solution is to be tested for \(\mathrm{Cl}^{-}\) ion by addition of 1 drop \((0.2 \mathrm{~mL})\) of \(0.10 \mathrm{M}\) \(\mathrm{AgNO}_{3}\). What is the minimum number of grams of \(\mathrm{Cl}^{-}\) that must be present for \(\mathrm{AgCl}(\mathrm{s})\) to form?

What is the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of water saturated with \(\mathrm{CO}_{2}\) at a partial pressure of \(1.10 \mathrm{~atm}\) ? The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-2} \mathrm{~mol} / \mathrm{L}-\mathrm{atm}\). The \(\mathrm{CO}_{2}\) is an acidic oxide, reacting with \(\mathrm{H}_{2} \mathrm{O}\) to form \(\mathrm{H}_{2} \mathrm{CO}_{3}\).

To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\) \(\left(K_{s p}=4 \times 10^{-10}\right)\) in \(1.0 \mathrm{~L}\) of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) because the solution will be quite basic.)
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