Question

Enter your answer in the provided box. An industrial chemist studying bleaching and sterilizing prepares a hypochlorite buffer using 0.450 M HClO and 0.450 M NaClO. (Ka for HClO = 2.9 × 10−8) Find the pH of 1.00 L of the solution after 0.040 mol of NaOH has been added.

Answer 1

Answer:

$\large \boxed{7.62}$

Explanation:

1. pH of original buffer

(a) Calculate pKₐ

$\text{p}K_{\text{a}} = -\log \left (K_{\text{a}} \right ) =-\log(2.9 \times 10^{-8}) = 7.54$

(b) Calculate the pH

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

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 7.54 +\log \left(\dfrac{0.450}{0.450}\right )\\\\& = & 7.54 + \log1.00 \\ & = & 7.54 + 0.00\\& = & 7.54\\\end{array}$

2. pH after adding strong base

(a) Find new composition of the buffer

The base reacts with the HA and forms A⁻

$\text{ Initial moles of HA} = \text{1.0 L} \times \dfrac{\text{0.450 mol }}{\text{1.00 L}} = \text{0.450 mol = 450 mmol}\\\\\text{ Initial moles of A}^{-} = \text{1.0 L} \times \dfrac{\text{0.450 mol }}{\text{1.00 L}} = \text{0.450 mol = 450 mmol}\\\\\text{Moles of OH ^{-} added} = \text{40 mmol}$

                 HA     +     OH⁻    ⟶     A⁻ + H₂O

I/mmol:    450            40              450

C/mmol:    -40           -40                 40

E/mmol:    410               0              490

(b) Find the new pH  

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 7.54 +\log \left(\dfrac{490}{410}\right )\\\\& = & 7.54 + \log1.195 \\& = & 7.54 +0.0774\\& = & \mathbf{7.62}\\\end{array}\\\text{The new pH is \large \boxed{\textbf{7.62}} }$