Question

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)

Answer 1

0.533 M KOH

Step-by-step 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