Question

A solution is prepared at 25 °C that is initially 0.23 M in propanoic acid, a weak acid with K, = 1.3 x 10^-5, and 0.42 M in sodium propanoate. Calculate the pH of the solution. Round your answer to 2 decimal places.​

Answer 1

Answer:

$\large \boxed{5.15}$

Explanation:

HA + H₂O ⇌ H₃O⁺ + A⁻

1. Calculate pKₐ

$\text{p}K_{\text{a}} = -\log \left (K_{\text{a}} \right ) =-\log(1.3 \times 10^{-5}) = 4.89$

2. Calculate the pH

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

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 4.89 +\log \left(\dfrac{0.42}{0.23}\right )\\\\& = & 4.89 + \log1.83 \\& = & 4.89 +0.261\\& = & 5.15\\\end{array}\\\text{The pH of the buffer is \large \boxed{\mathbf{5.15}} }$