A 10.0 mL of 0.121 M H2SO4 is neutralized by 17.1 mL of KOH solution according to the following balanced chemical reaction. H2SO4 (aq) + 2 KOH(aq) →K2SO4(aq) +

Question

A) Find the number of moles of KOH that reacted with H2SO4.
(that means moles of KOH that naturalized 10.0 mL of 0.121 M H2SO4)

B) What is the molarity of the KOH solution ?

Answer 1

H2SO4 (aq) + 2KOH(aq) ==> K2SO4(aq) + 2H2O <-Balanced equation

moles of H2SO4 present = 10.0 ml x 1 L/1000 ml x 0.121 mole/L = 0.00121 moles H2SO4

moles KOH needed = 0.00121 moles H2SO4 x 2 moles KOH/1 mole H2SO4 = 0.00242 moles KOH needed

Molarity KOH = moles/liter = 0.00242 moles/0.0171 liters = 0.142 M (to 3 significant figures)