Question

A 0.500-L sample of H₂SO₄ solution was analyzed by taking a 100.0mL aliquot and adding mL of . After the reaction occurred, an excess of ions remained in the solution. The excess base required mL of for neutralization. Calculate the molarity of the original sample of H₂SO₄. Sulfuric acid has two acidic hydrogens.

Answer 1

The molarity of the original sample of H2SO4 is 0.009289 M

Step-by-step explanation:

Step 1: The balanced equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Step 2: Calculating moles of NaOH

moles of NaOH = Molarity of NaOH * Volume of NaOH

moles of NaOH = 0.213 M * 50*10^-3 L

moles of NaOH = 0.01065 moles

Step 3: Calculating moles of HCL

HCl + NaOH → NaCl + H2O

13.21 mL of 0.103 M HCl

moles of HCl = molarity of HCl * Volume of HCl

moles of HCl = 0.103M * 13.21 *10^-3 L

moles of HCl =0.00136 moles

Step 4: Calculating moles of NaOH in excess

For 1 mole of HCl consumed there is 1 mole of NaOH needed to produce 1 mole of NaCl and 1 mole of H2O

For 0.00136 moles of HCl consumed, there is also 0.00136 moles of NaOH needed. Those are 0.00136 moles NaOH in excess.

Step 5: Calculating moles of NaOH reacted

0.01065 moles - 0.00136 moles = 0.009289 moles of NaOH reacted

Step 6: Calculating moles of H2SO4

There is 2 moles of NaOH consumed per 1 mole of H2SO4 consumed.

For 0.009289 moles of NaOH reacted, there is needed 0.009289/2 moles = 0.0046445 moles of H2SO4

Step 7: Calculating molarity of H2SO4

Molarity of H2SO4 = moles of H2SO4 / volume of H2SO4

Molarity of H2SO4 = 0.0046445 / 0.5L = 0.009289 M

The molarity of the original sample of H2SO4 is 0.009289 M