For the titration of 50.0 mL of 0.020 M aqueous salicylic acid with 0.020 M KOH, the pH of the solution after the addition of 55.0 mL of KOH is 10.98.
The reaction between salicylic acid and KOH is:
C₇H₆O₃(aq) + KOH(aq) ⇄ C₇H₅O₃⁻K⁺(aq) + H₂O(l)
The number of moles of salicylic acid and KOH before the reaction is:
[tex] n_{{C_{7}H_{6}O_{3}}_{i}} = 0.020 mol/L*0.050 L = 1.0 \cdot 10^{-3} \:mol [/tex]
[tex] n_{{KOH}_{i}} = 0.020 mol/L*0.055 L = 1.1 \cdot 10^{-3} \:mol [/tex]
After the addition of KOH, all the acid will react with KOH, and the number of moles of KOH remaining in the solution will be:
[tex] n_{{KOH}_{f}} = n_{{KOH}_{i}} - n_{{C_{7}H_{6}O_{3}}_{i}} = 1.1 \cdot 10^{-3} \:mol - 1.0 \cdot 10^{-3} \:mol = 1.0 \cdot 10^{-4} \:moles [/tex]
We need to find the concentration of this base with the total volume of solution:
[tex] C_{KOH} = \frac{n_{{KOH}_{f}}}{V_{t}} = \frac{1.0 \cdot 10^{-4} \:moles}{0.050 L + 0.055 L} = 9.52 \cdot 10^{-4} mol/L [/tex]
Now, with this concentration we can find the pH as follows:
[tex] pH = 14 - pOH [/tex]
[tex] pH = 14 - (-log[OH^{-}]) [/tex]
[tex]pH = 14 + log(9.52 \cdot 10^{-4}) = 10.98[/tex]
Therefore, the pH of the solution is 10.98.
You can learn more about pH here:
- https://brainly.com/question/491373?referrer=searchResults
- https://brainly.com/question/1525823?referrer=searchResults
I hope it helps you!
