# Question 156ba

Apr 25, 2017

pH = 5.13

#### Explanation:

We will use the $\text{Henderson-Hasselbach}$ equation to find the pH of this buffer solution

$\textcolor{w h i t e}{a a a a a a a a}$color(magenta)(pH = pKa + log[("conjugate base")/("weak acid"]]

$\text{Step 1}$
Find the equilibrium concentrations of the $\text{acetic acid}$(weak acid) and its conjugate base, $\text{sodium acetate}$. Find moles of each solute from their initial volume and initial concentration. Use total volume as $0.034 \text{ L}$.

$\textcolor{\mathmr{and} a n \ge}{\text{Acetic acid:}}$

• (0.010" L") * (1.0" M") = 0.01" moles"

• (0.01" moles")/(0.034" L") = 0.29" M"

$\textcolor{g r e e n}{\text{Sodium acetate:}}$

• (0.024" L") * (1.0" M") = 0.024" moles"

• (0.024" moles")/(0.034" L") =0.71" M"

$\text{Step 2}$
Look up the ${K}_{a}$ of $\text{acetic acid}$ from your chemistry book or the Web.
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$${K}_{a} = 1.8 \cdot {10}^{-} 5$
(source:preparatorychemistry.com/Bishop_weak_acid_Equilibrium.htm)

To find the $p K a$ of acetic acid, take the negative log of the ${K}_{a}$

• $p K a = - \log \left(1.8 \cdot {10}^{-} 5\right)$
• $p K a = 4.74$

$\text{Step 3}$
Plugin and solve

• $p H = p K a + \log \left[\left(\text{conjugate base")/("acid}\right)\right]$

• pH = 4.74 + log[(0.71 cancelM)/(0.29 cancelM")]#

• $p H = 4.74 + \log \left[2.45\right]$

• $p H = 4.74 + 0.39$

• $p H = 5.13$

You can check your answer by seeing that you clearly have more than $\text{2x}$ the amount of the $\text{conjugate base}$ than the $\text{weak acid}$ so your answer should reflect a higher $\text{pH}$.