Answer to Question #157556 in Organic Chemistry for Zofia

Question

What is the pH when 28.00 mL of 0.200 M potassium hydroxide (KOH) have been added to 50.00 mL of 0.100 M hydrochloric acid (HCl) solution?

Answer 1

$HCl + KOH \to KCl + H_2O$

number of moles of KOH = 28/1000 × 0.200 = 0.0056 moles.

number of moles of HCl = 50/1000 × 0.100 = 0.0050 moles.

Therefore, KOH is in excess and 0.0006 moles of it will be left after a complete neutralisation.

The total volume of the solution after the reaction is 28.00mL + 50.00mL = 78.00mL

$KOH \rightleftarrows K^+ + OH^-$

since 1 mole of KOH produces 1 mole of hydroxyl ion,

the number of moles of hydroxyl ion is 0.0006 moles

[OH^-] = 0.0006/0.0078 = 0.077M

pOH = -log(0.077) = 1.11

pH + pOH = 14

pH = 14 - 1.11 = 12.89