D36.1 pH and Ratio of Conjugate Base to Conjugate Acid

Jia Zhou; John Moore; Etienne Garand

D36.1 pH and Ratio of Conjugate Base to Conjugate Acid

Activity: equilibrium concentrations 1

Typical white vinegar contains 50. g of acetic acid (CH₃COOH) per 1.0 L solution, which makes it a 0.83 M aqueous acetic acid solution. The acid-base reaction between acetic acid and water is

with a K_a of 1.8 × 10^-5 at 25 °C.

If you have a beaker of vinegar at 25 °C, how much of each reaction species (CH₃COOH, CH₃COO¯, H₃O⁺) is present in the solution?

A. First, let’s calculate concentration of H₃O⁺. In other words, what is the pH of this 0.83 M acetic acid solution?

Do the work in your notebook, then left-click here for an explanation.

The acid dissociation constant expression for acetic acid is:

$K_a = \dfrac{[\text{CH}_3\text{COO}^{-}]_e[\text{H}_3\text{O}^{+}]_e}{[\text{CH}_3\text{COOH}]_e}$

Constructing an ICE table gives:

	CH₃COOH(aq)	+	H₂O(ℓ)	⇌	CH₃COO¯(aq)	+	H₃O⁺(aq)
Initial concentration (M)	0.83		–		0		1.0 × 10^-7
Change in concentration (M)	–x		–		+x		+x
Equilibrium concentration (M)	0.83 – x		–		x		1.0 × 10^-7 + x

Plugging the equilibrium concentrations into the K_a expression above gives:

$1.8 \times 10^{-5} = \dfrac{(x)(1.0 \times 10^{-7} + x)}{(0.83 - x)}$

Solving for x gives:

x = 3.8 × 10^-3 M

Therefore:

[H₃O⁺] = (1.0 × 10^-7 + x) M = ( 1.0 × 10^-7 + 3.8 × 10^-3) M = 3.8 × 10^-3 M

pH = -log ([H₃O⁺]) = -log (3.8 × 10^-3) = 2.42

B. The ICE table above also provides information about the concentrations of CH₃COOH and CH₃COO¯. We can use those values to calculate the percentage of acetic acid existing as the conjugate base (CH₃COO¯) in this solution. (This is quite similar to percent dissociation.)

Do the work in your notebook, then left-click here for an explanation.

From the ICE table above:

[CH₃COO¯]_e = x = 3.8 × 10^-3 M

Therefore:

$\%\;\text{CH}_3\text{COO}^{-} = \dfrac{[\text{CH}_3\text{COO}^{-}]_e}{[\text{CH}_3\text{COOH}]_0} \times 100\% = \dfrac{3.8 \times 10^{-3}\;M}{0.83\;M} \times 100\% = 0.46\%$

0.46% of the acetic acid in this 0.83 M solution exist as the conjugate base giving rise to pH = 2.42.

Imagine if we are able to follow around a single acetic acid molecule in this dynamic equilibrium: it is surrounded by water molecules, forming and breaking hydrogen-bonds as it moves around in the solution. Occasionally, a hydrogen-bond interaction leads to the transfer of a H⁺ to a water molecule, and the acetic acid becomes an acetate anion. As it continues traversing around in the solution as an acetate, it will encounter a H₃O⁺, leading to a H⁺ transferring back, and it is again in the acetic acid form. Following this acetic acid molecule around long enough, and we will find it in the CH₃COOH form 99.54% of the time, and in the CH₃COO¯ form 0.46% of the time.

Therefore, at any given moment in time, we will find 0.46% of the acetic acid in the CH₃COO¯ form in this vinegar solution.

Like any other system at equilibrium, when conditions are changed and the system is no longer at equilibrium, the system will react to achieve new equilibrium concentrations.

Activity: equilibrium concentrations 2

Because the concentration of H₃O⁺ is part of this equilibrium, changing the pH of this solution will perturb the equilibrium.

Predict qualitatively what will happen to the solution pH and [CH₃COO¯]_e when NaOH is added to the solution.

Write in your notebook, then left-click here for an explanation.

The added OH¯ will react with H₃O⁺ present to form H₂O, this effectively lowers [H₃O⁺], i.e., the pH of the solution increases when OH¯ is added.

In response to this perturbation, the reaction

CH₃COOH(aq) + H₂O(ℓ) ⇌ CH₃COO¯(aq) + H₃O⁺(aq)

will proceed in the forward direction (towards products) to each a new equilibrium. Therefore, [CH₃COO¯]_e will increase when OH¯ is added.

If we again imagine following around a single acetic acid molecule: when it is in the CH₃COO¯ form, it will encounter H₃O⁺ less frequently because the added OH¯ have reacted away many of the H₃O⁺. Therefore, it will spend more time in the CH₃COO¯ form, leading to an equilibrium where there are more CH₃COO¯ present at any given moment.

For any acid-base equilibrium, changing the solution pH will change the fraction of acid that exists in its conjugate form (this applies equally to a base and its conjugate acid). For example, we can plot the percentage of acetic acid and acetate anion in a solution as a function of pH:

As the graph above shows, there’s a point where the conjugate pair is at a 50:50 ratio. The pH at this point equals the acid’s pK_a value. This relationship can be derived from the K_a expression:

$K_a = \dfrac{[\text{H}_3\text{O}^{+}]_e[\text{A}^{-}]_e}{[\text{HA}]_e}$

Rearranging the variables gives:

$[\text{H}_3\text{O}^{+}]_e = K_a\;\times\;\dfrac{[\text{HA}]_e}{[\text{A}^{-}]_e}$

Taking the negative logarithm of both sides gives:

$\begin{array}{rcl} -\text{log}[\text{H}_3\text{O}^{+}]_e &=& -\text{log}\;K_a\;-\;\text{log}\dfrac{[\text{HA}]_e}{[\text{A}^{-}]_e} \\[1 em] \text{pH} &=& \text{p}K_a\;-\;\text{log}\dfrac{[\text{HA}]_e}{[\text{A}^{-}]_e} \\[1 em] \text{pH} &=& \text{p}K_a\;+\;\text{log}\dfrac{[\text{A}^{-}]_e}{[\text{HA}]_e} \end{array}$

Therefore, when [HA]_e = [A¯]_e, $\text{log}\dfrac{[\text{A}^{-}]_e}{[\text{HA}]_e}$ = log(1) = 0, and pH = pK_a, as the graph above shows.

This equation can also be rearranged to show that the ratio of conjugate base to conjugate acid is determined by the solution pH:

$\dfrac{[\text{A}^{-}]_e}{[\text{HA}]_e}\;=\;10^{(\text{p}\text{H}-\text{p}K_{\text{a}})}$

If a solution’s pH is two units above the acid’s pK_a value, 99% of the acid is in its conjugate base form. Conversely, if the pH is two units below the pK_a, 1% of the acid is in its conjugate base form. In other words, raising the solution pH shifts the equilibrium toward the conjugate base and lowering the solution pH shifts the equilibrium toward the acid.

raise pH →

$\text{HA(aq)}\ \ + \ \ \text{H}_2\text{O}(\ell)\ \ \rightleftharpoons \ \ \text{A}^-\text{(aq)}\ \ +\ \ \text{H}_3\text{O}^+\text{(aq})$

← lower pH

The equation verifies quantitatively the prediction of Le Chatelier’s principle that adding strong base would react away hydronium ions, shifting the equilibrium toward products.

Exercise: Acid/base Ratio and pH

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Chem 109 Fall 2024 Copyright © by Jia Zhou; John Moore; and Etienne Garand is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.