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# 21.22: Calculating pH of Salt Solutions

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Many enjoy a cool dip in a swimming pool on a hot day, but 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 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.

##### Example $$\PageIndex{1}$$

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

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

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 $$1.4 \times 10^{-11}$$.

###### 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

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*}\nonumber

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

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

\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*}\nonumber

###### Step 3: Think about your result.

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:

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

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 $$2.00 \: \text{M}$$.

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

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

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

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*}\nonumber

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

## Summary

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