General Chemistry chapter 5

Question

Suppose 0.450L of 0.420M H2SO4 is mixed with 0.400L of 0.250M KOH. What concentration of sulfuric acid remains after neutralization?

Answer 1

Write the correctly balanced equation for the reaction:

H2SO4 + 2KOH ==> K2SO4 + 2H2O (NOTE the ratio of H2SO4 to KOH is 1 to 2)

moles KOH present = 0.400 L x 0.250 mol/L = 0.100 moles

moles H2SO4 present = 0.450 L x 0.420 mol/L = 0.189 moles

moles H2SO4 neutralized by KOH = 0.100 moles KOH x 1 mol H2SO4 / 2 moles KOH = 0.0500 moles H2SO4 reacted (neutralized)

moles H2SO4 remaining = 0.189 mols - 0.05 mols = 0.139 mols

Concentration of H2SO4 remaining = mols / total volume and total volume is now 0.450 L + 0.400 L = 0.850L

H2SO4 = 0.139 mols / 0.850 L = 0.164 M