15.00 mL of 0.425 M H2SO4 solution is required to completely react with (neutralize) 23.9 mL

of KOH solution.

What is the

molarity of the KOH solution?

2 KOH (aq) + H2SO4 (aq) → K2SO4 (aq) + 2 H20 (1)

0.533 M KOH

Explanation:

2 KOH (aq) + H2SO4 (aq) → K2SO4 (aq) + 2 H20 (l}

from reaction 2 mol 1 mol

x 15*10^ (-3) * 0.425 mol

15.00 mL = 15*10^ (-3) L H2SO4 solution

15*10^ (-3) L*0.425 mol/L = 15*10^ (-3) * 0.425 mol H2SO4

x = 2*15*10^ (-3) * 0.425/1 = 2*15*10^ (-3) * 0.425 mol KOH

23.9mL = 23.9 * 10^ (-3) L KOH

M (KOH) - molarity KOH

M (KOH) * V (KOH solution) = M (KOH) * 23.9 * 10^ (-3) L KOH - is moles KOH

2*15*10^ (-3) * 0.425 mol H2SO4 = M (KOH) * 23.9 * 10^ (-3) L KOH

M (KOH) = 2*15*10^ (-3) * 0.425 / (23.9 * 10^ (-3)) = 0.533 M