Step 1. Calculate the moles of sulfuric acid in Beaker A
The density of the solution is 1.015 g/mL.
#"Mass of solution" = 300 color(red)(cancel(color(black)("mL solution"))) × "1.015 g solution"/(1 color(red)(cancel(color(black)("mL solution")))) = "304 g solution"#
#"Mass of H"_2"SO"_4 = 304 color(red)(cancel(color(black)("g solution"))) × ("2.5 g H"_2"SO"_4)/(100 color(red)(cancel(color(black)("g solution")))) = "7.61 g H"_2"SO"4#
#"Moles of H"_2"SO"_4 = 7.61 color(red)(cancel(color(black)("g H"_2"SO"_4))) × ("1 mol H"_2"SO"_4)/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) = "0.0776 mol H"_2"SO"_4#
Step 2. Calculate the moles of sulfuric acid in Beaker B
#" Moles of H"_2"SO"_4 = 0.350 color(red)(cancel(color(black)("L solution"))) × ("0.25 mol H"_2"SO"_4)/(1 color(red)(cancel(color(black)("L solution")))) = "0.0875 mol H"_2"SO"_4#
Step 3. Calculate the total moles of sulfuric acid after adding Beaker A to
Beaker B
#"Total moles = (0.0776 + 0.0875) mol = 0.165 mol"#
Step 4. Calculate the moles of barium nitrate in Beaker C
#"Moles of Ba"("NO"_3)_2 = 0.350 color(red)(cancel(color(black)("L solution"))) × ("0.02 mol Ba"("NO"_3)_2) /(1 color(red)(cancel(color(black)("L solution")))) = "0.0070 mol Ba"("NO"_3)_2#
Step 5. Calculate #Q_text(sp)# for the mixture formed by adding Beaker C to Beaker B
#"Total volume = (300 + 350 + 350) mL = 1000 mL = 1.00 L"#
#["Ba"^"2+"] = "0.0070 mol"/"1.00 L" = "0.0070 mol/L"#
#["SO"_4^"2-"] = "0.165 mol"/"1.00 L" = "0.165 mol/L"#
#"BaSO"_4 ⇌ "Ba"^"2+" + "SO"_4^"2-"; K_text(sp) = 9.9 × 10^"-11"#
#Q_text(sp) = ["Ba"^"2+"]["SO"_4^"2-"] = 0.0070 × 0.165 = 1.2 × 10^"-3"#
#Q_text(sp) > K_text(sp)#
∴ A white precipitate of #"BaSO"_4# will form.
