ABC - Weak Bases and pH (Lesson)

Weak Bases and pH

The previous lesson explored the properties of weak acids as well as how to perform calculations related to pH and K_a. This lesson will explore the category of substances known as weak bases. Weak bases are characterized as substances that, when dissolved in water, readily accept a proton thereby producing hydroxide (OH^-) ions in solution. A generic weak base equilibrium and equilibrium expression are shown below.

B (aq) + H₂O (l) HB⁺ (aq) + OH^- (aq)

$K_{b} = \frac{[H B^{+}] [O H^{-}]}{[B]}$

WeakBases

On first inspection, it may be thought that instead of pH, the pOH of weak base solutions must be calculated. While this is certainly possible, it is far more common to express the basicity of a solution by calculating its pH, keeping in mind that the range of 7-14 on the pH scale is considered basic.

As an example, consider a 0.20 M solution of the weak base NH₃ (K_b = 1.8 x 10^-5). How would the pH of this solution be calculated? The first step, as with weak acids, is to write the chemical equilibrium and equilibrium expression:

NH₃ (aq) + H₂O (l) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ NH₄⁺ (aq) + OH^- (aq)

$LaTeX: K_b=\frac{\left\lbrack NH_4^+\right\rbrack\left\lbrack OH^{-^{}}\right\rbrack}{\left\lbrack NH_3\right\rbrack}=1.8\times10^{-5}$ $K_b=\frac{\left\lbrack NH_4^+\right\rbrack\left\lbrack OH^{-^{}}\right\rbrack}{\left\lbrack NH_3\right\rbrack}=1.8\times10^{-5}$

Once again, an ICE table proves helpful in organizing information to solve for the required values:

ICE TABLE
	NH₃ (aq)	H₂O (l)	NH₄⁺ (aq)	OH^- (aq)
Initial	0.20 M	--	0	0
Change	- x	--	+ x	+ x
Equilibrium	0.20 - x	--	x	x

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

$LaTeX: K_b=\frac{\left\lbrack NH_4^+\right\rbrack\left\lbrack OH^{-^{}}\right\rbrack}{\left\lbrack NH_3\right\rbrack}=\frac{\left(x\right)\left(x\right)}{0.20-x}=1.8\times10^{-5}$ $K_b=\frac{\left\lbrack NH_4^+\right\rbrack\left\lbrack OH^{-^{}}\right\rbrack}{\left\lbrack NH_3\right\rbrack}=\frac{\left(x\right)\left(x\right)}{0.20-x}=1.8\times10^{-5}$

Again a situation arises where, when the expression is expanded, a quadratic equation is encountered. However, as before, the assumption can be made that the value of x is very small compared to 0.20 in this case. Omitting the x in the denominator avoids the use of the quadratic formula and results in the following:

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

$LaTeX: x^2=\left(.20\right)\left(1.8\times10^{-5}\right)=3.6\times10^{-6}$ $x^2=\left(.20\right)\left(1.8\times10^{-5}\right)=3.6\times10^{-6}$

$LaTeX: x=\sqrt{3.6\times10^{-5}}=1.9\times10^{-3}$ $x=\sqrt{3.6\times10^{-5}}=1.9\times10^{-3}$

Notice in this case that x does not refer to the concentration of hydronium at equilibrium but rather the concentration of hydroxide. Because of this, the pH cannot be calculated directly and pOH must be calculated first:

pOH = -log [OH^-]

pOH = - log (1.9 x 10^-3)

pOH = 2.72

pH + pOH = 14.00

pH = 14.00 -2.72

pH = 11.28

You Try It!

In the following self-assessment activity, calculate the OH^- concentration and pH. Click on the plus sign to check your answer!

[CC BY 4.0] UNLESS OTHERWISE NOTED | IMAGES: LICENSED AND USED ACCORDING TO TERMS OF SUBSCRIPTION