Question

If 50mL of a 5.43x10^-3M solution of perchloric acid in water is mixed with 10mL of a 1.34x10^-2M solution of potassium hydroxide in water, what is the pH of the mixture?

*** I know the answer is 2.64, but I need someone to please explain how to get the answer.

Answer 1

V1 = 50 ml

M1 = 5.43*10-3 HClO4

V2 = 1.34*10^-2

M2 = 1.34*10^-2 KOH

Since there are strong bases and acids, we just need to calculate the remaining [H+] or [OH-] ions

moels of HClO4 = M1 * V1 = 50 ml * 5.43*10^-3 = 0.27 mmol

moles of KOH = M2 *V2 = 10 ml * 1.32*10^-2 = 0.13 mmol

0.27 milimoles of H+ and 0.13 moles of OH- react to form 0.13 milimoles of water and an excess of 0.14 moels of H+

since strong acid:

HClO4 -> H+ and ClO4-

Concentration of [H+] = moles [H+] / volume

[H+] = 0.14 mmol / 60 ml = 0.0023

NOTE: the total volume has changed from 50 ml to 10 ml = 60 ml is now the total volume

Now pH = -log[H+] = -log 0.0023 = 2.63

pH = 2.63