Answer:
pH = 8.0
Explanation:
First, we have to calculate the moles of NaOH.
![35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol](https://tex.z-dn.net/?f=35.8%20%5Ctimes%2010%5E%7B-3%7DL.%5Cfrac%7B0.020mol%7D%7BL%7D%20%3D7.2%5Ctimes%2010%5E%7B-4%7Dmol)
Let's consider the balanced equation.
C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O
The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.
The concentration of C₂H₃O₃Na is:
![\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M](https://tex.z-dn.net/?f=%5Cfrac%7B7.2%5Ctimes%2010%5E%7B-4%7Dmol%7D%7B60.8%20%5Ctimes%2010%5E%7B-3%7DL%7D%20%3D0.012M)
C₂H₃O₃Na dissociates according to the following equation:
C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)
C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.
C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻
If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:
pKa + pKb = 14
pKb = 14 -3.9 = 10.1
10.1 = -log Kb
Kb = 7.9 × 10⁻¹¹
We can calculate [OH⁻] using the following expression:
[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>
[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M
Now, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0
pH + pOH = 14
pH = 14 - pOH = 14 - 6.0 = 8.0