Question

The pH of a solution prepared by mixing 45 mL of 0.183 M KOH and 35 mL of 0.145 M HCl is

Answer 1

HCl is strong acid and dissociates completely to give H+ and Cl- ions

KOH is strong base and dissociates completely to give K+ and OH- ions

HCl moles = M x V = ( 0.145 ) x ( 35/1000) = 0.005075 = H+ moles

KOH moles = 0.183 x ( 45/1000) = 0.008235 = OH- moles

now OH- moles left after neutralisation = 0.008235-0.005075 = 0.00316

total vol = 45+35 = 80 ml = 0.08 L

[OH-] = ( 0.00316/0.08) = 0.0395

pOH = -log [OH-] = -log ( 0.0395) = 1.4

pH = 14-pOH = 14-1.4 = 12.6