Answer:
[tex]\rm 3.56 \times 10^{-5}\, mol/L[/tex]
Explanation:
HA + H₂O ⇌ H₃O⁺ + A⁻; pKₐ = 4.75
A solution of a weak acid and a salt is a buffer.
(a) 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 )\\\\& = &4.75 +\log \left(\dfrac{0.05}{0.10}\right )\\\\& = & 4.75 + \log0.50 \\& = & 4.75 - 0.30\\& = &4.45 \\\end{array}[/tex]
(b) Calculate [H₃O⁺]
[tex]\rm [H_{3}O^{+}] = 10^{-pH} = 10^{-4.45} = 3.56 \times 10^{-5}\, mol/L[/tex]
