Answer:
[tex]\large \boxed{7.62}[/tex]
Explanation:
1. pH of original buffer
(a) Calculate pKₐ
[tex]\text{p}K_{\text{a}} = -\log \left (K_{\text{a}} \right ) =-\log(2.9 \times 10^{-8}) = 7.54[/tex]
(b) Calculate the pH
We can use the Henderson-Hasselbalch equation to get the pH.
[tex]\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}[/tex]
2. pH after adding strong base
(a) Find new composition of the buffer
The base reacts with the HA and forms A⁻
[tex]\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}[/tex]
HA + OH⁻ ⟶ A⁻ + H₂O
I/mmol: 450 40 450
C/mmol: -40 -40 40
E/mmol: 410 0 490
(b) Find the new pH
[tex]\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 $\large \boxed{\textbf{7.62}}$}[/tex]
