Question

A 10.0mL 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) + 2 H2O(l)

A) Find the number of moles of KOH that neutralized 10.0mL of 0.121 M H2SO4.

B) What is the molarity of the KOH solution?

Answer 1

moles Given Molasity = Volume in litre im = 10-3L) Volume of thson = 10.0mL = 1.00x 102 L Molarity of Hasoy = 0.121 M of H₂SO4 = (0.121M) ( 100 x 10-2L). of H₂SO4 = 0.00121 mol a moles Moles H₂ Sousa + 2KOH lan Koutan + 2H₂0 W imol amol mol amol (stoichiometric, coefficiente) To neutralize, I I'mol H₂SO4 needs 2 mol KOH = 0.00121 mol H, sou needs 260.0021 mol) =0.00242 mol Kon So, Moles of KOH needed = 0.00242 mol ) I Given W We know volume of KOH Volume of koH moles of kot = 17,1 mL = 1711x10 3 2 = 0.00242 mo! Molarity of koH = o. 0.00242 mol T710 10 3 L Mdarity of tot = 0.142) (3 sig fy) KOH = 0.142M