Question

At 25°C, a saturated solution of benzoic acid (Ka 6.4 × 10-⁵) has a pH of 2.80. Calculate the = water solubility of benzoic acid in moles per liter.

Answer 1

The water solubility of benzoic acid is approximately
`\(0.00156 \, \text{M}\)` at 25°C.

To find the water solubility of benzoic acid
`(\(C_6H_5COOH\))` at 25°C, we'll follow these steps:

1. Determine the concentration of
`\(H^+\)` ions:

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

`\[ [H^+] = 10^{-\text{pH}} \]`

`\[ [H^+] = 10^(-2.80) \]`

2. Set up the equilibrium expression using the given
`\(K_a\)` value:

`\[ K_a = ([C_6H_5COO^-][H^+])/([C_6H_5COOH]) \]`

3. Let x be the concentration of
`\(H^+\)` ions and
`(\(C_6H_5COO^-\))` at equilibrium. The initial concentration of benzoic acid
`\([C_6H_5COOH]_0\)` is the solubility we are trying to find.

`\[ 6.4 * 10^(-5) = (x^2)/([C_6H_5COOH]_0) \]`

4. Solve for \([C_6H_5COOH]_0\):

`\[ [C_6H_5COOH]_0 = (x^2)/(6.4 * 10^(-5)) \]`

Since
`\(x = [H^+]\),` we can substitute
`\(10^(-2.80)\) for \(x\)`.

`\[ [C_6H_5COOH]_0 = ((10^(-2.80))^2)/(6.4 * 10^(-5)) \]`

`\[ [C_6H_5COOH]_0 = (10^(-5.60))/(6.4 * 10^(-5)) \]`

`\[ [C_6H_5COOH]_0 = (1)/(640) \]`

`\[ [C_6H_5COOH]_0 = 0.0015625 \, \text{M} \]`

So, the water solubility of benzoic acid is approximately
`\(0.00156 \, \text{M}\)` at 25°C.