21.22: Calculating pH of Salt Solutions

We all enjoy a cool dip in a swimming pool on a hot day, but we may not realize the work needed to keep that water safe and healthy. The ideal pH for a swimming pool is around 7.2. The pH will change as a result of many factors. Adjustment can be accomplished with different chemicals depending on the tested pH. High pH can be lowered with liquid $$\ce{HCl}$$ (unsafe material) or sodium bisulfate. The bisulfate anion is a weak acid and can dissociate partially in solution. To increase pH, use sodium carbonate. The carbonate anion forms an equilibrium with protons that result in some formation of carbon dioxide.

Calculating pH of Salt Solutions

It is often helpful to be able to predict the effect a salt solution will have on the pH of a certain solution. Knowledge of the relevant acidity or basicity constants allows us to carry out the necessary calculations.

Example 21.22.1

If we dissolve $$\ce{NaF}$$ in water, we get the following equilibrium:

$\ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)$

The pH of the resulting solution can be determined if the $$K_\text{b}$$ of the fluoride ion is known. $$20.0 \: \text{g}$$ of sodium fluoride is dissolved in enough water to make $$500.0 \: \text{mL}$$ of solution. Calculate the pH of the solution. The $$K_\text{b}$$ of the fluoride ion is $$1.4 \times 10^{-11}$$.

Solution:

Step 1: List the known values and plan the problem.

Known

• Mass $$\ce{NaF} = 20.0 \: \text{g}$$
• Molar mass $$\ce{NaF} = 41.99 \: \text{g/mol}$$
• Volume solution $$= 0.5000 \: \text{L}$$
• $$K_\text{b}$$ of $$\ce{F^-} = 1.4 \times 10^{-11}$$

Unknown

• pH of solution $$= ?$$

The molarity of the $$\ce{F^-}$$ solution can be calculated from the mass, molar mass, and solution volume. Since $$\ce{NaF}$$ completely dissociates, the molarity of the $$\ce{NaF}$$ is equal to the molarity of the $$\ce{F^-}$$ ion. An ICE table (below) can be used to calculate the concentration of $$\ce{OH^-}$$ produced and then the pH of the solution.

Step 2: Solve.

\begin{align} 20.0 \: \cancel{\text{g} \: \ce{NaF}} \times \frac{1 \: \cancel{\text{mol} \: \ce{NaF}}}{41.99 \: \cancel{\text{g} \: \ce{NaF}}} \times \frac{1 \: \text{mol} \: \ce{F^-}}{1 \: \cancel{\text{mol} \: \ce{NaF}}} &= 0.476 \: \text{mol} \: \ce{F^-} \\ \frac{0.476 \: \text{mol} \: \ce{F^-}}{0.5000 \: \text{L}} &= 0.953 \: \text{M} \: \ce{F^-} \end{align}

$\text{Hydrolysis equation:} \: \: \: \ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)$

$\begin{array}{l|ccc} & \ce{F^-} & \ce{HF} & \ce{OH^-} \\ \hline \text{Initial} & 0.953 & 0 & 0 \\ \text{Change} & -x & +x & +x \\ \text{Equilibrium} & 0.953 - x & x & x \end{array}$

\begin{align} K_\text{b} &= 1.4 \times 10^{-11} = \frac{\left( x \right) \left( x \right)}{0.953 - x} = \frac{x^2}{0.953 - x} \approx \frac{x^2}{0.953} \\ x &= \left[ \ce{OH^-} \right] = \sqrt{1.4 \times 10^{-11} \left( 0.953 \right)} = 3.65 \times 10^{-6} \: \text{M} \\ \text{pOH} &= -\text{log} \left( 3.65 \times 10^{-6} \right) = 5.44 \\ \text{pH} &= 14 - 5.44 = 8.56 \end{align}

The solution is slightly basic due to the hydrolysis of the fluoride ion.

Salts That Form Acidic Solutions

When the ammonium ion dissolves in water, the following equilibrium exists:

$\ce{NH_4^+} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{NH_3} \left( aq \right)$

The production of hydronium ions causes the resulting solution to be acidic. The pH of a solution of ammonium chloride can be found in a very similar way to the sodium fluoride solution in the previous example. However, since the ammonium chloride is acting as an acid, it is necessary to know the $$K_\text{a}$$ of $$\ce{NH_4^+}$$, which is $$5.6 \times 10^{-10}$$. We will find the pH of a $$2.00 \: \text{M}$$ solution of $$\ce{NH_4Cl}$$. Because the $$\ce{NH_4Cl}$$ completely ionizes, the concentration of the ammonium ion is $$2.00 \: \text{M}$$.

$\ce{NH_4Cl} \left( s \right) \rightarrow \ce{NH_4^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)$

Again, an ICE table (below) is set up in order to solve for the concentration of the hydronium (or $$\ce{H^+}$$) ion produced.

$\begin{array}{l|ccc} & \ce{NH_4^+} & \ce{H^+} & \ce{NH_3} \\ \hline \text{Initial} & 2.00 & 0 & 0 \\ \text{Change} & -x & +x & +x \\ \text{Equilibrium} & 2.00 - x & x & x \end{array}$

Now substituting into the $$K_\text{a}$$ expression gives:

\begin{align} K_\text{a} &= 5.6 \times 10^{-10} = \frac{x^2}{2.00 - x} \approx \frac{x^2}{2.00} \\ x &= \left[ \ce{H^+} \right] = \sqrt{ 5.6 \times 10^{-10} \left( 2.00 \right)} = 3.3 \times 10^{-5} \: \text{M} \\ \text{pH} &= -\text{log} \left( 3.3 \times 10^{-5} \right) = 4.48 \end{align}

A salt produced from a strong acid and a weak base yields a solution that is acidic.

Summary

• Calculations to determine the pH of salt solutions are described.

Contributors

• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.