Answer:
[tex]\large \boxed{4.7}[/tex]
Explanation:
The equation for the buffer equilibrium is
HA + H₂O ⇌ H₃O⁺ + A⁻
1. Calculate the composition of the original buffer
[tex]\text{ Initial moles of HA} = \text{50 mL} \times \dfrac{\text{0.20 mmol }}{\text{1 mL}} = \text{10 mmol}\\\\\text{ Initial moles of A}^{-} = \text{50 mL} \times \dfrac{\text{0.20 mol }}{\text{1.00 L}} = \text{10 mmol}[/tex]
So you have 10 mmol of HA and 10 mmol of A⁻ in 100 mL of the buffer.
2. pH after adding NaOH
(a) Find new composition of the buffer
The base reacts with the HA and forms A⁻
[tex]\text{Moles of OH$^{-}$ added} = \text{5 mL } \times \dfrac{\text{0.10 mmol}}{\text{ 1 mL}} = \text{0.50 mmol}[/tex]
HA + H₂O ⇌ H₃O⁺ + A⁻
I/mmol: 10 10
C/mmol: -0.5 +0.5
E/mmol: 9.5 10.5
(b) Find the new pH
[tex]\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 4.7 +\log \left(\dfrac{10.5}{9.5}\right )\\\\& = & 4.7 + \log1.11 \\& = & 4.7 +0.043\\& = & \mathbf{4.7}\\\end{array}\\\text{The new pH is $\large \boxed{\textbf{4.7}}$}[/tex]
