Question

In the neutralization reaction,

H2SO4 + 2 KOH = 2 H2O + K2SO4

If there is 887 mL of 0.239 M H2SO4 reacting with 467 mL of a KOH solution with unknown concentration, what is the concentration of KOH?

Answer 1

Number of moles of H2SO4 = molarity * volume of solution in L

Number of moles of H2SO4 = 0.239 * 0.887 L = 0.212 mole

From the balanced equation we can say that

1 mole of H2SO4 requires 2 mole of KOH so

0.212 mole of H2SO4 will require

= 0.212 mole of H2SO4 *( 2 mole of KOH / 1 mole of H2SO4)

= 0.424 mole of KOH

Molarity of KOH = number of moles of KOH / volume of solution in L

Molarity of KOH = 0.424 mol / 0.467 L = 0.908 M

Therefore, concentration of KOH = 0.908 M