The composition of an aqueous solution is 0. I M acetic acid +0.2M sodium acetate (a) Calculate the pH of the solution (pk, of acetic acid is 4.74). 3 pts (b) What would the pH be if 2 ml of 10M NaOH was added to 1 liter of this solution? 4 pts. (c). By comparison, what would the pH be if the same amount of NaOH were added to IL of pure water?

[tex]\begin{array}{rcl}\text{pH}& = &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\& = &4.74 + \log\dfrac{0.2}{0.1}\\\\& = &4.74 + \log 2\\& = &4.74 + 0.30\\& = & \mathbf{5.04}\\\end{array}[/tex]

(b) pH after addition of base

[tex]\text{Moles of NaOH} = \text{0.002 L} \times \dfrac{\text{10 mol}}{\text{1 L }} = \text{0.02 mol}[/tex]

HA + H₂O ⇌ H₃O⁺ + A⁻

I/mol: 0.1 0 0.2

C/mol: -0.02 +x +0.02

E/mol: 0.08 x 0.22

[tex]\text{pH}& =4.74 + \log\dfrac{0.22}{0.08}\\\\& = & 4.74 + \log 2.75\\& = & 4.74 + 0.44\\& = & \mathbf{5.18}\\\end{array}[/tex]

(c) pH of water after addition of base

[tex]\text{[OH$^{-}$]} = \dfrac{\text{0.02 mol}}{\text{1.002 L}} = \text{0.02 mol/L}[/tex]

pOH} = -log0.02 = 1.70

pH = 14.00 - pOH = 14.00 - 1.70 = 12.30