Question

Calculate the percent ionization and the pH of each of the following solutions of benzoic acid, \(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.3 \times 10^{-5}\right)\) : (a) \(1.0 \mathrm{M}\) (b) \(0.10 \mathrm{M}\) (c) \(0.010 \mathrm{M}\)

Short answer

For 1.0 M benzoic acid: \text{pH}=2.10, \% ionization=0.794\%. For 0.10 M: \text{pH}=2.60, \% ionization=2.51\%. For 0.010 M: \text{pH}=3.10, \% ionization=8.00\%.

Step-by-step solution

- Write the ionization equation

Benzoic acid (\text{HC}_7\text{H}_5\text{O}_2) ionizes in water as follows:\[\text{HC}_7\text{H}_5\text{O}_2 \rightleftharpoons \text{H}^+ + \text{C}_7\text{H}_5\text{O}_2^-\]

- Write the expression for the acid dissociation constant, \(K_a\)

The expression for the dissociation constant \(K_a\) is given by:\[K_a = \frac{[\text{H}^+][\text{C}_7\text{H}_5\text{O}_2^-]}{[\text{HC}_7\text{H}_5\text{O}_2]}\]Given: \(K_a = 6.3 \times 10^{-5}\)

- Establish initial concentrations

Let the initial concentration of benzoic acid be represented as follows for each case:\[\begin{align*} \text{(a) } &[HC_7H_5O_2]=1.0 M ewline \text{(b) } &[HC_7H_5O_2]=0.10 M ewline \text{(c) } &[HC_7H_5O_2]=0.010 M \ \end{align*}\]

- Set up the ICE table

Create an ICE (Initial, Change, Equilibrium) table for the ionization:\[\begin{array}{c|c|c|c} & [\text{HC}_7\text{H}_5\text{O}_2] & [\text{H}^+] & [\text{C}_7\text{H}_5\text{O}_2^-] \ \text{Initial (I)} & C & 0 & 0 \ \text{Change (C)} & -x & +x & +x \ \text{Equilibrium (E)} & C-x & x & x \ \text{Expression} \end{array}\]

- Approximate if possible and solve for x

In this step, for strong enough initial concentrations where \(C-x\approx C\), approximate the equilibrium concentration and solve for x:\[6.3 \times 10^{-5} = \frac{x^2}{C-x} \approx \frac{x^2}{C}\]Solving for x,\[x \approx \sqrt{6.3 \times 10^{-5} C}\]

- Calculate percent ionization

Percent ionization is calculated as:\[\% \text{ ionization} = \left( \frac{x}{C} \right) \times 100\]

- Calculate the pH

pH is given by:\[\text{pH} = -\log[\text{H}^+]\]Since \([\text{H}^+]=x\), calculate the pH.

- Plug in values and compute for each concentration

Now calculate the specific values for each initial concentration:(a) \[\text{For } C = 1.0 M:\]\[x \approx \sqrt{6.3 \times 10^{-5} \times 1.0 M} = 7.94 \times 10^{-3} M\]\[\% \text{ ionization} = \left( \frac{7.94 \times 10^{-3}}{1.0} \right) \times 100 = 0.794\%\]\[\text{pH} = -\log(7.94 \times 10^{-3}) = 2.10\](b) \[\text{For } C = 0.10 M:\]\[x \approx \sqrt{6.3 \times 10^{-5} \times 0.10 M} = 2.51 \times 10^{-3} M\]\[\% \text{ ionization} = \left( \frac{2.51 \times 10^{-3}}{0.10} \right) \times 100 = 2.51\%\]\[\text{pH} = -\log(2.51 \times 10^{-3}) = 2.60\](c) \[\text{For } C = 0.010 M:\]\[x \approx \sqrt{6.3 \times 10^{-5} \times 0.010 M} = 8.00 \times 10^{-4} M\]\[\% \text{ ionization} = \left( \frac{8.00 \times 10^{-4}}{0.010} \right) \times 100 = 8.00\%\]\[\text{pH} = -\log(8.00 \times 10^{-4}) = 3.10\]

Key concepts

acid dissociation constant (Ka)

The acid dissociation constant, represented as \(K_a\), is used to quantify the strength of a weak acid. It indicates how well the acid can donate a proton (H+) in an aqueous solution.
The general formula to express \(K_a\) is:

\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]

Where:

• \([\text{H}^+]\) is the concentration of hydrogen ions< \li>
• \([\text{A}^-]\) is the concentration of the conjugate base
• \([\text{HA}]\) is the concentration of the undissociated acid

In the given exercise, benzoic acid is considered a weak acid with a \(K_a\) value of \(6.3 \times 10^{-5}\), which indicates its ionization ability in water.
To solve problems involving \(K_a\), we often need to set up an ICE table, solve for the equilibrium concentrations, and then use these values to calculate the percent ionization and pH.

ICE table

An ICE table helps organize and simplify the process of solving equilibrium problems. 'ICE' stands for Initial, Change, and Equilibrium - the three stages we need to consider.

1. **Initial:** This represents the initial concentrations before the reaction starts.
2. **Change:** This shows how the concentrations change as the reaction proceeds.
3. **Equilibrium:** This gives the final concentrations at equilibrium.

For example, when setting up the ICE table for benzoic acid's ionization:

• **Initial:** \([HC_7H_5O_2] = C\), \([\text{H}^+] = 0\), \([\text{C}_7H_5O_2^-] = 0\)
• **Change:** \([\text{HC}_7H_5O_2]\) reduces by \(x\), \([\text{H}^+]\) increases by \(x\), \([\text{C}_7H_5O_2^-]\) increases by \(x\)
• **Equilibrium:** \([HC_7H_5O_2] = C - x\), \([\text{H}^+] = x\), \([\text{C}_7H_5O_2^-] = x\)

Using these, we create the expression:
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{x^2}{C - x} \]
If \(C\) is much larger than \(x\), we can approximate \(C - x \approx C\), simplifying our calculations.

pH calculation

pH is a measure of the acidity or basicity of a solution. It is defined using the negative logarithm of the hydrogen ion concentration (\text{H}^+). Mathematically, this is expressed as:

\[ \text{pH} = -\text{log} [\text{H}^+] \]

For benzoic acid solutions, after determining the equilibrium concentration of \([\text{H}^+]\) (which is \(x\) from the ICE table), we can directly find the pH.

For example, if \([\text{H}^+] = 7.94 \times 10^{-3} \text{M}\):

\[ \text{pH} = -\text{log} (7.94 \times 10^{-3}) = 2.10 \]

With these steps, solving for pH becomes straightforward after calculating the hydrogen ion concentration from the \(K_a\) expression.

Additionally, we calculate percent ionization to understand how much of the benzoic acid has ionized:

\[ \text{percent ionization} = \frac{x}{C} \times 100 \]
For a concentration of \(1.0 \text{M}\), the ionization would be about \(0.794\text{%}\), combining both pH and ionization measurement allows a thorough evaluation of the acid's dissociation.