What will be the pH of a solution containing 100 mol of potassium hydroxide and 0.05 mol of sulfuric acid in 100 cm3 of solution?
The reaction can be represented as follows:
2KOH + H2SO4 = K2SO4 + 2H2O
From here, after the reaction, 1.00 mol - (2 × 0.05 mol) = 0.9 mol of KOH are left in the solution.
The molar concentration of KOH equals:
M(KOH) = n(KOH) / V(KOH) = 0.9 mol / 100 cm3 = 0.9 mol / 0.1 L = 9 M
From here, the pOH of the solution:
pOH = −log[OH−] = −log = 0.95
Finally, pH equals:
pH = 14 - pOH = 14 - 0.95 = 13.05