16.6: Weak Acids - Chemistry LibreTexts

Example \(\PageIndex{8}\): Equilibrium Concentrations in a Solution of a Weak Acid

Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the \(\ce{HSO4-}\) ion, a weak acid. What is the pH of a 0.50-M solution of \(\ce{HSO4-}\)?

\[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \nonumber \]

Solution

We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of \(\ce{HSO4-}\) so that we can use \(\ce{[H3O+]}\) to determine the pH. As in the previous examples, we can approach the solution by the following steps:

1. Determine \(x\) and equilibrium concentrations. This table shows the changes and concentrations:

2. Solve for \(x\) and the concentrations.

As we begin solving for \(x\), we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of \(x\).

At equilibrium:

\[K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+][SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x} \nonumber \]

If we assume that x is small and approximate (0.50 − x) as 0.50, we find:

\[x=7.7×10^{−2} \nonumber \]

When we check the assumption, we confirm:

\[\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \nonumber \]

which for this system is

\[\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \nonumber \]

The value of \(x\) is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find \(x\).

The equation:

\[K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x}\nonumber \]

gives

\[6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \nonumber \]

or

\[x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0 \nonumber \]

This equation can be solved using the quadratic formula. For an equation of the form

\[ax^{2+} + bx + c=0, \nonumber \]

\(x\) is given by the quadratic equation:

\[x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \nonumber \]

In this problem, \(a = 1\), \(b = 1.2 × 10^{−3}\), and \(c = −6.0 × 10^{−3}\).

Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:

\[x=7.2×10^{−2} \nonumber \]

Now determine the hydronium ion concentration and the pH:

\[\begin{align*} \ce{[H3O+]} &=~0+x=0+7.2×10^{−2}\:M \\[4pt] &=7.2×10^{−2}\:M \end{align*} \nonumber \]

The pH of this solution is:

\[\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \nonumber \]