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:

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I hope it helps you!

First by getting moles of salicylic acid:

moles of salicylic acid = molarity * volume 

                                     = 0.02 * 0.05 L

                                     = 0.001 mol

then moles of base KOH = molarity * volume 

                                           = 0.02 * 0.055L

                                           = 0.0011

when total volume = 0.05 + 0.055 = 0.105 L

[salisalic acid] = moles / total volume

                        = 0.001 / 0.105

                        = 0.0095

[KOH] = moles / total volume

           = 0.0011 / 0.105

           = 0.01

by using H-H equation, we can get the PH:

 PH = Pka + ㏒[salt/acid]

by substitution:

PH = 2.97 + ㏒[0.01 / 0.0095]

      = 2.99 

It's a

Electrons are emitted