Question

The pH of a solution prepared by mixing 65 mL of 0.183 M KOH and 45 mL of 0.145 M HCl is ______.

Answer 1

mo ofmoles of KOH = molarity of KOH x volume of KOH in liters

= 0.183 M x 0.065L = 0.012 moles

no ofmoles of HCl = 0.145M x 0.045 L = 0.0065 moles

now write the balanced equation

KOH + HCl ----> KCl + H2O

from this equation it is clear that one mole of HCl required one mole of KOH

here HCl moles are less compare to KOH

0.0065 moles of HCl consume the 0.0065 moles of KOH

so no of moles of KOH remains 0.012 - 0.0065 = 0.0055 moles of KOH remains remains

to tak volume = 65 + 45 = 110 mL =0.110 L

molarity of KOH remains in the solution = no of moles remains / total volume

= 0.0055 / 0.110L

= 0.05M

since KOH is strong base [KOH] = [OH^-] = 0.05 M

pOH = -log[0.05M]

pOH = 1.30

pH + pOH = 14

from this

pH = 14-pOH = 14-1.30 = 12.7