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# Calculating pH of Salt Solutions

## Describes the ICE method and gives examples.

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Calculating pH of Salt Solutions

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License: CC BY-NC 3.0

Keeping things safe and healthy

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 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 results 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.

#### Sample Problem: Salt Hydrolysis

If we dissolve NaF in water, we get the following equilibrium:

The pH of the resulting solution can be determined if the  \begin{align*}K_b\end{align*} of the fluoride ion is known.

20.0 g of sodium fluoride is dissolve in enough water to make 500.0 mL of solution. Calculate the pH of the solution. The  \begin{align*}K_b\end{align*} of the fluoride ion is 1.4 × 10 −11 .

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

Known

• mass NaF = 20.0 g
• molar mass NaF = 41.99 g/mol
• volume solution = 0.500 L
• \begin{align*}K_b\end{align*} of F - = 1.4 × 10 −11

Unknown

• pH of solution = ?

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

Step 2: Solve.

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

\begin{align*}\text{Hydrolysis equation:} \qquad \text{F}^-(aq)+\text{H}_2\text{O}(l)\rightleftarrows \text{HF}(aq)+\text{OH}^-(aq)\end{align*}

 Concentrations [F - ] [HF] [OH - ] Initial 0.953 0 0 Change \begin{align*}-x\end{align*} \begin{align*}+x\end{align*} \begin{align*}+x\end{align*} Equilibrium \begin{align*}0.953 - x\end{align*} \begin{align*}x\end{align*} \begin{align*}x\end{align*}

\begin{align*}K_b &=1.4\times 10^{-11}=\frac{(x)(x)}{0.953-x}=\frac{x^2}{0.953-x}\approx \frac{x^2}{0.953} \\ x &=[\text{OH}^-]=\sqrt{1.4\times 10^{-11}(0.953)}=3.65\times 10^{-6} \ \text{M} \\ \text{pOH} &=-\log (3.65\times 10^{-6})=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:

\begin{align*}\text{NH}_4^+(aq)+\text{H}_2\text{O}(l)\rightleftarrows \text{H}_3\text{O}^+(aq)+\text{NH}_3(aq)\end{align*}

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 Sample Problem 21.7. However, since the ammonium chloride is acting as an acid, it is necessary to know the \begin{align*}K_a \end{align*} of NH 4 + , which is 5.6 × 10 −10 . We will find the pH of a 2.00 M solution of NH 4 Cl. Because the NH 4 Cl completely ionizes, the concentration of the ammonium ion is 2.00 M.

\begin{align*}\text{NH}_4\text{Cl}(s)\rightarrow \text{NH}_4^+(aq)+\text{Cl} ^-(aq)\end{align*}

Again, an ICE Table ( below ) is set up in order to solve for the concentration of the hydronium (or H + ) ion produced.

 Concentrations [NH 4 + ] [H + ] [NH 3 ] Initial 2.00 0 0 Change \begin{align*}-x\end{align*} \begin{align*}+x\end{align*} \begin{align*}+x\end{align*} Equilibrium \begin{align*}2.00 - x\end{align*} \begin{align*}x\end{align*} \begin{align*}x\end{align*}

Now substituting into the \begin{align*}K_a\end{align*} expression gives:

\begin{align*}K_a &=5.6 \times 10^{-10}=\frac{x^2}{2.00-x} \approx \frac{x^2}{2.00} \\ x &=[\text{H}^+]=\sqrt{5.6 \times 10^{-10}(2.00)}=3.3 \times 10^{-5} \ \text{M} \\ \text{pH} &=- \log (3.3 \times 10^{-5})=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 pH of salt solutions are described.

#### Practice

Work the problems at the link below:

#### Review

Questions

1. In the first example, how do we know that we can ignore  \begin{align*}x\end{align*} when determining [F - ]?
2. In example two, how do we know the ammonium ion concentration?
3. Could we write the equilibrium in example two as \begin{align*}\text{NH}_4^+ \rightleftarrows \text{H}^+ + \text{NH}_3\end{align*} ?