For the titration of 50.0 ml of 0.020 m aqueous salicylic acid with 0.020 m koh (aq), calculate the ph after the addition of 55.0 ml of the base. for salycylic acid, pka = 2.97.'

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:

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:

We need to find the concentration of this base with the total volume of solution:

Now, with this concentration we can find the pH as follows:

Therefore, the pH of the solution is 10.98.

You can learn more about pH here:

I hope it helps you!

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