ABC - Weak Acids and pH (Lesson)

Weak Acids and pH

In the previous lesson, the pH of solutions of strong acids and bases was determined. These calculations were relatively simple because strong acids, by definition, dissociate completely when placed in aqueous solution, making the concentration of the acid equal to the concentration of hydronium that is produced in the solution. However, what about weak acids? How can weak acids be adequately described?

Weak Acids

Since strong acids dissociate 100% when placed in water, it follows that weak acids would therefore dissociate by less than 100%. This is indeed true and furthermore, they typically dissociate by far less than 100%. Instead of completely dissociating, weak acids establish a dynamic equilibrium in much the same way as all other equilibria that have been discussed thus far. If the abbreviation HA is used to represent a weak acid, then an equilibrium can be described as follows:

HA (aq) + H₂O (l) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ H₃O⁺ (aq) + A^- (aq)

An equilibrium expression describing this scenario can also be written:

$LaTeX: K_a=\frac{\left\lbrack H_{3_{}}O^+\right\rbrack\left\lbrack A^-\right\rbrack}{\left\lbrack HA\right\rbrack}$ $K_a=\frac{\left\lbrack H_{3_{}}O^+\right\rbrack\left\lbrack A^-\right\rbrack}{\left\lbrack HA\right\rbrack}$

Notice that this equilibrium also receives a special designation as K_a (the acid dissociation constant) which indicates that this is a weak acid equilibrium.

Calculating K_a from pH

As with any other type, the equilibrium constant can be calculated if the concentrations of all substances after equilibrium has been achieved are known. This idea, in conjunction with concepts related to pH, allows for equilibrium constants to be calculated.

Take for example a 0.20 M solution of acetic acid that has a pH of 2.72. The first step in determining the K_a for this acid would be to write out the equilibrium and the equilibrium expression:

CH₃COOH (aq) + H₂O (l) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ CH₃COO^- (aq) + H₃O⁺ (aq)

$LaTeX: K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack CH_3COO^-\right\rbrack}{\lbrack CH_3COOH]}$ $K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack CH_3COO^-\right\rbrack}{\lbrack CH_3COOH]}$

Using the pH provided, the [H₃O⁺] at equilibrium can be determined:

pH = -log[H₃O⁺] = 2.72

log[H₃O⁺] = -2.72

[H₃O⁺] = 10^-2.72 = 1.9 x 10^-3 M

At this point, an ICE table may prove helpful to organize the information, but it must be noted that this is an example of a heterogeneous equilibrium due to the fact that water is a liquid. Because only aqueous solutions and gases appear in equilibrium expressions, water has been omitted from both the ICE table and the equilibrium expression.

Table
	CH₃COOH (aq)	H₂O (l)	CH₃COO^- (aq)	H₃O⁺ (aq)
Initial	0.20 M	--	0	0
Change	- 1.9 x 10^-3 M	--	+ 1.9 x 10^-3 M	+ 1.9 x 10^-3 M
Equilibrium	(0.20 - 1.9 x 10^-3) M	--	1.9 x 10^-3 M	1.9 x 10^-3 M

Using the values from the Equilibrium line of the table and substituting them into the equilibrium expression from above, the following value for K_a is obtained:

$LaTeX: K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack CH_3COO^-\right\rbrack}{\left\lbrack CH_3COOH\right\rbrack}$ $K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack CH_3COO^-\right\rbrack}{\left\lbrack CH_3COOH\right\rbrack}$

$LaTeX: K_a=\frac{\left(1.9\times10^{-3}\right)\left(1.9\times10^{-3}\right)}{\left(0.20-1.9\times10^{-3}\right)}=1.8\times10^{-5}$ $K_a=\frac{\left(1.9\times10^{-3}\right)\left(1.9\times10^{-3}\right)}{\left(0.20-1.9\times10^{-3}\right)}=1.8\times10^{-5}$

Percent Ionization

Because K_a is simply an equilibrium constant for a specific type of chemical equilibrium, it can be utilized to compare acid strength among weak acids. For example, when comparing two weak acids, the substance with the highest K_a value (indicating a greater concentration of hydronium at equilibrium) will have the greatest acid strength. An additional way to compare acid strength is through the use of percent ionization. As the name implies, percent ionization is a measure of the extent to which an acid is dissociated and can be calculated using the relationship shown below:

$LaTeX: \%ionization=\frac{\left\lbrack H_3O^+\right\rbrack_{eq}}{\left\lbrack HA\right\rbrack_0}\times100$ $\%ionization=\frac{\left\lbrack H_3O^+\right\rbrack_{eq}}{\left\lbrack HA\right\rbrack_0}\times100$

Using the values from above, the percent ionization of a 0.20 M acetic acid solution can be determined:

$LaTeX: \%ionization=\frac{\left\lbrack H_3O^+\right\rbrack_{eq}}{\left\lbrack HA\right\rbrack_0}=\frac{\left(1.9\times10^{-3}\right)}{\left(0.20\right)}\times100=0.95\%$ $\%ionization=\frac{\left\lbrack H_3O^+\right\rbrack_{eq}}{\left\lbrack HA\right\rbrack_0}=\frac{\left(1.9\times10^{-3}\right)}{\left(0.20\right)}\times100=0.95\%$

Using K_a to Calculate pH

A far more common type of calculation is to utilize a provided K_a value to calculate the pH of a solution. Assume that a solution of formic acid acid (HCOOH) has a concentration of 0.13 M and a K_a value of 1.8 x 10^-4. The steps below illustrate the process necessary to calculate the pH of this solution at equilibrium. As before, the first step should be to write the equilibrium and equilibrium expression that describes this scenario:

HCOOH (aq) + H₂O (l) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ HCOO^- (aq) + H₃O⁺ (aq)

$LaTeX: K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack HCOO^-\right\rbrack}{\left\lbrack HCOOH\right\rbrack}=1.8\times10^{-4}$ $K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack HCOO^-\right\rbrack}{\left\lbrack HCOOH\right\rbrack}=1.8\times10^{-4}$

Once again an ICE table could prove beneficial to help organize the information, but notice a significant difference. In this case, we have no information regarding the concentration of either product present at equilibrium. In this instance, it must be stated simply that the concentration of formic acid decreases by some amount "x" while the product concentrations increase by some amount "x" as the system approaches equilibrium:

Table
	HCOOH (aq)	H₂O (l)	HCOO^- (aq)	H₃O⁺ (aq)
Initial	0.13 M	--	0	0
Change	- x	--	+ x	+ x
Equilibrium	0.13 - x	--	x	x

Substituting the values on the Equilibrium line into the equilibrium expression shown above yields the following:

$LaTeX: K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack HCOO^-\right\rbrack}{\left\lbrack HCOOH\right\rbrack}=\frac{\left(x\right)\left(x\right)}{0.13-x}=1.8\times10^{-4}$ $K_a=\frac{\left\lbrack H_3O^+\right\rbrack\left\lbrack HCOO^-\right\rbrack}{\left\lbrack HCOOH\right\rbrack}=\frac{\left(x\right)\left(x\right)}{0.13-x}=1.8\times10^{-4}$

It can be observed that this expression, when expanded out, would yield an equation that would require the quadratic formula to solve. However, in the case of weak acid equilibria, a key property can be taken advantage of in these cases. The equilibrium constant, Ka, for weak acids is typically very small indicating that the concentrations of the products at equilibrium are also small. These concentrations are so small in fact that the "x" in the denominator in the expression above can be neglected. The only caveat to this assumption is that the percent ionization must be 5% or less. If the percent ionization is greater than 5%, then the "x" value cannot be ignored and the quadratic equation must be used to solve the problem. Ignoring the "x" in the above equation simplifies the equation as shown:

$LaTeX: \frac{\left(x\right)\left(x\right)}{0.13}=1.8\times10^{-4}$ $\frac{\left(x\right)\left(x\right)}{0.13}=1.8\times10^{-4}$

$LaTeX: x^2=\left(0.13\right)\left(1.8\times10^{-4}\right)=2.3\times10^{-5}$ $x^2=\left(0.13\right)\left(1.8\times10^{-4}\right)=2.3\times10^{-5}$

$LaTeX: x=\sqrt{2.3\times10^{-5}}=4.8\times10^{-3}$ $x=\sqrt{2.3\times10^{-5}}=4.8\times10^{-3}$

After solving for the value of x and recognizing that x represents both the concentration of the conjugate base and the concentration of hydronium present at equilibrium, this value can then be used to solve for the pH of the solution;

pH = -log[H₃O⁺]

pH = -log (4.8 x 10^-3)

pH = 2.32

It must be noted that this value was calculated after making the assumption that the x in the denominator was negligible. Was this a valid assumption though? In order to verify this the % ionization must be calculated to ensure that it is 5% or less:

$LaTeX: \%ionization=\frac{\left\lbrack H_3O^+\right\rbrack_{eq}}{\left\lbrack HA\right\rbrack_0}=\frac{\left(4.8\times10^{-3}\right)}{\left(0.13\right)}\times100=3.7\%$ $\%ionization=\frac{\left\lbrack H_3O^+\right\rbrack_{eq}}{\left\lbrack HA\right\rbrack_0}=\frac{\left(4.8\times10^{-3}\right)}{\left(0.13\right)}\times100=3.7\%$

As the calculation reveals, the % ionization is less than 5% and the assumption that x was negligible was a valid assumption.

You Try It!

In the following self-assessment activity, complete the requested calculation. Click on the plus sign to check your answer!

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