KEY2CHEM

Relationship of pH and Acidity

The $$pH$$ ($$pH = -log[\require{mhchem}\ce{H3O+}]$$) of an aqueous solution is determined by the identity and concentration of the substance that is dissolved in water. For acid-base systems, pH characterizes the relative availability of protons. The ability of an acid, HA, to donate a proton in water is measured by its equilibrium constant, $$K_a$$:

$$\require{mhchem}\ce{HA(aq) + H2O(l) <=>H3O+(aq) + A^{-}(aq)} \;\;\;\; K_a$$

The $$pK_a$$ of an acid ($$pK_a = - log K_a$$) is another measure of acid strength. A smaller $$pK_a$$ corresponds to a larger $$K_a$$, which corresponds to greater proton transfer by the acid.

Example 1.

A weak acid, $$\require{mhchem}\ce{HX}$$, has a $$pK_a$$ of $$4.76$$. If the $$pH$$ of the solution is $$3.08$$, which species is mostly present in solution?

A. $$\require{mhchem}\ce{HX}$$

B. $$\require{mhchem}\ce{X-}$$

C. equal mixture of $$\require{mhchem}\ce{HX}$$ and $$\require{mhchem}\ce{X-}$$

Solution

A. $$\require{mhchem}\ce{HX}$$

Since $$pH < pK_a$$, most of the acid is in its protonated form.

Example 2.

Which solution will have the lowest $$pH$$? The $$pK_a$$ of each acid is listed.

A. $$0.1 \text{ M } \require{mhchem}\ce{HCl};\; pK_a = -7.00$$

B. $$0.1 \text{ M } \require{mhchem}\ce{HF}; \;pK_a = 3.47$$

C. $$0.1 \text{ M } \require{mhchem}\ce{CH3COOH};\; pK_a = 4.76$$

Solution

A. $$0.1 \text{ M } \require{mhchem}\ce{HCl}; \;pK_a = -7.00$$

The acid with the lowest $$pK_a$$ (which corresponds to the greatest proton transfer when dissolved in water) will have the lowest $$pH$$.

Example 3.

A solution contains $$0.25\text{ M }$$ of weak acid $$\require{mhchem}\ce{HA}$$ and $$1.0 \text{ M}$$ of its conjugate base $$\require{mhchem}\ce{A-}$$. The $$pK_a$$ of $$\require{mhchem}\ce{HA}$$ is $$5.22$$. What is true about the $$pH$$ of the solution?

A. $$pH = 5.22$$

B. $$pH > 5.22$$

C. $$pH < 5.22$$

Solution

B. $$pH > 5.22$$

Most of the solution is composed of the deprotonated form of the acid (its conjugate base, $$\require{mhchem}\ce{A-}$$). As such, the solution’s $$pH$$ is determined by the predominant form, which yields a solution with $$pH > pK_a$$ of the acid. This could also be proved numerically with the Henderson-Hasselbalch equation.

$$pH = pK_a + log \frac{\require{mhchem}\ce{A-}}{\require{mhchem}\ce{HA}}$$

$$pH = 5.22 + log \frac{1.0\text{ M}}{0.25\text{ M}}$$

$$pH = 5.22 + log 4$$

$$pH = 5.82$$