If 18.5 mL of 1.9 10-4 M HBr is added to 34.5 mL of 7.6 10-4 M KOH, what is the pH of the solution?

Answer 1

18.5*1.9*10^-4 = 35.15x 10^-4 millimol

34.5*7.6*10^-4 = 262.2 x 10^-4 millimol

Moles of KOH left = 227.05 x 10^-4 millimol

Volume = 18.5 + 34.5 = 53 ml

Thus molarity = 227.05/53 x 10^-4 M = 4.284x10^-4 M

pH = 14+log(4.284x10^-4) = 10.63

Answer 2

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