Answer:
[tex]\large \boxed{4.7}[/tex]
Explanation:
HA + H₂O ⇌ H₃O⁺ + A⁻
(a) Calculate pKₐ
[tex]\text{p}K_{\text{a}} = -\log \left (K_{\text{a}} \right ) =-\log(1.8 \times 10^{-5}) = 4.74[/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 )\\\\& = &4.74 +\log \left(\dfrac{0.01}{0.01}\right )\\\\& = & 4.74 + \log1.0 \\& = & 4.74 +0.0\\& = &4.7 \\\end{array}\\\text{The pH is $\large \boxed{\textbf{4.7}}$}[/tex]
