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.

The molarity of the original sample of H2SO4 is 0.009289 M

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