Question

What is the pH of a buffer that was prepared by adding 3.96 g of sodium benzoate, NaC7H5O2, to 1.00 L of 0.0100 M benzoic acid, HC7H5O2? Assume that there is no change in volume. The Ka for benzoic acid is 6.3 × 10–5. View Available Hint(s) What is the pH of a buffer that was prepared by adding 3.96 g of sodium benzoate, NaC7H5O2, to 1.00 L of 0.0100 M benzoic acid, HC7H5O2? Assume that there is no change in volume. The Ka for benzoic acid is 6.3 × 10–5. 4.33 3.76 0.439 4.64

Answer 1

`\large \boxed{4.64}`

Step-by-step explanation:

The equation for the buffer is

HC₇H₅O₂ + H₂O ⇌ H₃O⁺ + C₇H₅O₂⁻; Kₐ =6.3 × 10⁻⁵

Let's rewrite it as

HA + H₂O ⇌ H₃O⁺ + A⁻

1. Moles of sodium benzoate

`\text{Moles} = \text{3.96 g} * \frac{\text{1 mol}}{\text{144.10 g}} = \text{0.027 49 mol}`

2. Concentration of benzoate ion

`[\text{A}^(-)] = \frac{\text{0.027 49 mol}}{\text{1 L}} = \text{0.027 49 mol/L }`

3. Calculate pKₐ

`\text{p}K_{\text{a}} = -\log(6.3 * 10^(-5)) = 4.20`

4. Calculate the pH

We can use the Henderson-Hasselbalch equation to get the pH.

`\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log\frac{[\text{A}^(-)]}{\text{[HA]}}\\\\& = & 4.20 +\log(0.02749)/(0.0100)\\\\& = & 4.20 + \log2.749 \\& = & 4.20 + 0.4390\\& = & 4.64\\\end{array}\\\text{The pH of the buffer is $\large \boxed{\mathbf{4.64}}$}`