Define convert back to molarity

Reference no: EM13175673

For the titration of 50 mL of 0.020 M aqueous salicylic acid with 0.020 M KOH(aq), calculate the pH after the addition of 55 mL of KOH(aq). For salycylic acid, pK_a = 2.97. From what I know, after converting to moles and adding new KOH, I thought you were suppose to subtract the moles of acid(.001mol) from moles of KOH(.0021), then convert back to molarity. And since it is a strong acid, you would just take the -log(KOH or OH-) then subtract this number from 14. From this I get 11.85, however the answer is 10.98. Can somebody tell me where I'm going wrong in detail?

Reference no: EM13175673

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