Question

30.0 mL of 0.512 M sulfuric acid was added to 25.0 mL of 0.666 M sodium

hydroxide.

a. Write the balanced chemical equation.

b. Determine the molarity of either acid or base remaining after the reaction.

c. Find the molarity if a 50.00 mL aliquot of this solution is diluted to 250. mL.

Answer 1

Given,

Concentration of H₂SO₄ solution = 0.512 M

Volume of H₂SO₄ solution = 30.0 mL x ( 1L/1000 mL) = 0.03 L

Concentration of NaOH solution = 0.666 M

Volume of NaOH solution = 25.0 mL x ( 1L/1000 mL) = 0.025 L

a) The balanced chemical reaction between H₂SO₄ and NaOH is,

H₂SO₄(aq) + 2NaOH(aq) Na₂SO₄(aq) + 2H₂O(l)

b) Now, Calculating the number of moles of sulfuric acid and sodium hydroxide given,

= 0.512 M x 0.030 L = 0.0154 mol H₂SO₄

Similarly,

= 0.666 M x 0.025 L = 0.01665 mol NaOH

Now, Calculating the number of moles of NaOH required to react completely with the given moles of H₂SO₄,

= 0.0154 mol H₂SO₄ x ( 2 mol NaOH / 1 mol H₂SO₄)

= 0.03072 mol NaOH is required to react completely with the given moles of H₂SO₄.

Also, Calculating the number of moles of H₂SO₄ required to react completely with the given moles of NaOH,

= 0.01665 mol NaOH x ( 1 mol H₂SO₄ / 2 mol NaOH)

= 0.008325 mol H₂SO₄ is required to react completely with the given moles of NaOH.

Thus, NaOH is limiting reactant and H₂SO₄ is an excess reactant.

Now, calculating the excess reactant,

= Given moles of H₂SO₄ - Moles of H₂SO₄ reacted.

= 0.0154 mol H₂SO₄ - 0.008325 mol

= 0.007035 mol of H₂SO₄

Now,

the total volume = 0.03 L + 0.025 L

total volume = 0.055 L

New molarity(M) = Number of moles / Volume of solution

New molarity(M) = 0.007035 mol of H₂SO₄ / 0.055 L solution

New molarity(M) = 0.128 M

c) We know,

Molarity of solution = 0.128 M

Volume of the aliquot of this solution taken(V) = 50.00 mL x ( 1L /1000 mL) = 0.05 L

Calculating the number of moles

= 0.128 M x 0.05 L

= 0.006395 moles

Now, dilution volume = 250 mL /1000 mL = 0.250 L

Molarity = Moles / L of solution

Molarity = 0.006395 moles / 0.250 L of solution

Molarity = 0.0256 M