This is really a two-part question:
- How much of the "H"_2"SO"_4 is neutralized?
- What is the concentration of the remaining "H"_2"SO"_4?
1. How much of the "H"_2"SO"_4 is neutralized?
This is a volume-moles stoichiometry problem.
The steps are:
a. Write the balanced equation.
b. Use the molarity of "KOH" to convert volume of "KOH" to moles of "KOH".
c. Use the molar ratio from the balanced equation to convert moles of "KOH" to moles of "H"_2"SO"_4 that reacted.
d. Use the molarity to convert the initial volume of "H"_2"SO"_4 to moles of "H"_2"SO"_4.
e. Calculate the moles of excess "H"_2"SO"_4.
a. Write the balanced equation:
"2KOH" + "H"_2"SO"_4 → "K"_2"SO"_4 + "2H"_2"O"
b. Calculate the moles of "KOH".
"Moles of KOH" = 0.800 color(red)(cancel(color(black)("L KOH"))) × "0.280 mol KOH"/(1 color(red)(cancel(color(black)("L KOH")))) = "0.224 mol KOH"
c. Calculate the moles of "H"_2"SO"_4 that reacted
The molar ratio of "H"_2"SO"_4:"KOH" is "1 mol H"_2"SO"_4:"2 mol KOH".
∴ "Moles of H"_2"SO"_4 = 0.224 color(red)(cancel(color(black)("mol KOH"))) × (1 "mol H"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol KOH")))) = "0.112 mol H"_2"SO"_4
2. What is the concentration of the remaining "H"_2"SO"_4?
a. Calculate the initial moles of "H"_2"SO"_4.
"Moles of H"_2"SO"_4 = 0.850 color(red)(cancel(color(black)("L H"_2"SO"_4))) × ("0.490 mol H"_2"SO"_4)/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) = "0.4165 mol"
e. Calculate the moles of "H"_2"SO"_4 remaining
"Moles remaining" = "initial moles – moles reacted" = "0.4165 mol – 0.112 mol" = "0.3045 mol"
e. Calculate the molarity of the excess "H"_2"SO"_4
"Molarity" = "moles"/"litres"
"Litres" = "Volume of KOH + Volume of H"_2"SO"_4 = "0.800 L + 0.850 L" = "1.650 L"
∴ "Molarity" = "0.3045 mol"/"1.650 L" = "0.185 mol/L"
