Question

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

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.

Homework Answers

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

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