0.850 L of 0.490 M #H_2SO_4# is mixed with 0.800 L of 0.280 M #KOH#. What concentration of #H_2SO_4# remains after neutralization?

Answer 1

The concentration of #"H"_2"SO"_4# after partial neutralization is 0.185 mol/L

This is really a two-part question:

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"#
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Answer 2

To find the concentration of H₂SO₄ remaining after neutralization, calculate the moles of H₂SO₄ and KOH, then determine which reactant is limiting. The limiting reactant will be fully consumed, and the excess reactant will determine the concentration of the remaining acid. The balanced chemical equation for the reaction is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

The moles of H₂SO₄ and KOH are: moles H₂SO₄ = (volume H₂SO₄) × (molarity H₂SO₄) moles KOH = (volume KOH) × (molarity KOH)

Next, determine the limiting reactant by comparing the moles of H₂SO₄ and KOH.

Once you find the limiting reactant, use stoichiometry to determine the moles of H₂SO₄ that reacted.

Finally, use the remaining volume of H₂SO₄ and the moles of H₂SO₄ remaining to calculate the concentration of H₂SO₄.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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