Finding the equilibrium Constant from pH

A 0.100 M weak acid (HA) solution has a pH of 4.25. Find K $_{a}$ for the acid.

Question

Finding the equilibrium Constant from pH

A 0.100 M weak acid (HA) solution has a pH of 4.25. Find K[latex]_{a}[/latex] for the acid.

Accepted Answer

Use the given pH to find the equilibrium concentration of [ [latex]H_{3}O^{+}[/latex] ]. Then write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing all known concentrations.

pH = -log [ [latex]H_{3}O^{+}[/latex] ]
4.25 = -log [ [latex]H_{3}O^{+}[/latex] ]
[ [latex]H_{3}O^{+}[/latex] ] = 5.6 × 10[latex]^{-5}[/latex] M
[latex]HA(aq) + H_{2}O(l) \xrightleftharpoons[]{} H_{3}O^{-}(aq) + A^{-}(aq)[/latex]

	[HA]	[latex][H_{3}O^{+}][/latex]	[A[latex]^{-}[/latex]]
Initial	0.100	≈0.00	0.00
Change
Equil		5.6×10[latex]^{-5}[/latex]

Use the equilibrium concentration of[latex]H_{3}O^{+}[/latex] and the stoichiom etry of the reaction to predict the changes and equilibrium concentration for all species. For most weak acids, the initial and equilibrium concentrations of the weak acid (HA) are equal because the amount that ionizes is usually very small compared to the initial concentration.

[latex]HA(aq) + H_{2}O(l) \xrightleftharpoons[]{} H_{3}O^{-}(aq) + A^{-}(aq)[/latex]

	[HA]	[latex][H_{3}O^{+}][/latex]	[A[latex]^{-}[/latex]]
Initial	0.100	≈0.00	0.00
Change	-5.6×10[latex]^{-5}[/latex]	+5.6×10[latex]^{-5}[/latex]	+5.6×10[latex]^{-5}[/latex]
Equil	(0.100-×10[latex]^{-5}[/latex])≈0.100	5.6×10[latex]^{-5}[/latex]	5.6×10[latex]^{-5}[/latex]

Substitute the equilibrium concentrations into the expression for K[latex]_{a}[/latex] and calculate its value.

[latex]K_{a} = \frac{[H_{3}O^{+}][A^{-}]}{[HA]}[/latex]
=[latex] \frac{(5.6 imes 10^{-5})(5.6 imes 10^{-5})}{0.100}[/latex]
= 3.1 × 10[latex]^{-8}[/latex]

Question 15.8: Finding the equilibrium Constant from pH A 0.100 M weak acid...

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