Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. neutralization reaction: Suppose 0.250 L of 0.450 M H2SO4 is mixed with 0.200 L of 0.240 M KOH . What concentration of sulfuric acid remains after neutralization

0.197 M

Explanation:

The reaction equation is:

H2SO4(aq) +2KOH(aq) > K2SO4(aq) + 2H2O(l)

number of moles of H2SO4 = 0.25 L * 0.45 M = 0.1125 moles

number of moles of KOH = 0.2 L * 0.24 M = 0.048 moles

since H2SO4 is the reactant in excess;

2 moles of KOH reacts with 1 mole of H2SO4

0.048 moles of KOH reacts with 0.048 * 1/2 = 0.024 moles of H2SO4

Amount of excess H2SO4 left unreacted = 0.1125 - 0.024 = 0.0885 moles

Total volume = 0.25 L+ 0.2 L = 0.45 L

concentration of H2SO4 = 0.0885/0.45 = 0.197 M