Question

Find pH and pOH of a 3:1 mixture of 0.0125 M H2SO4 and 0.0125 M KOH

Ans: pH = 1.200 and pOH = 12.800

Answer 1

Given 0.0125 M H2SO4 and 0.0125 M KOH with ratio of 3: 1

Mole ratio of H2SO4 and KOH for 1.0 L solution

For 1.0 L moles = molariy

Moles of H2SO4 = 3 x 0.0125 mol = 0.0375 mol

moles of H+ from H2SO4 = 2 x moles of H2So4 = 2 x 0.0375 mol = 0.075 mol H+

Moles of KOH = 1 x 0.0125 mol = 0.0125 H+ mol

moles of OH- from KOH = 0.0125 mol OH-

Net moles of H+ = 0.075 - 0.0125 mol = 0.0625 mol

Now conc of H+ = mole / volume = 0.0625 mol / 1.0 L = 0.0625 M

pH of solution = -log [H+] = -log 0.0625 = 1.204 = 1.20

pH + pOH = 14

pOH = 14 -pH = 14 - 1.20 = 12.8