[Titration101]

Hi! I have a problem for a Titration Reaction, to do with uncertainties. The base/titre was NaOH, the acid was H₂SO₄ the volumes were both known for H₂SO₄, 10.0mL ± 0.10mL and for the NaOH 10.73 ± 0.1mL. The concentration of the NaOH was also known, 0.0963M (no uncertainty). The question asks that I find the concentration value of H₂SO₄. I then found this value to be 0.0517M (through C x V of NaOH to find the moles, then x 0.5 (mole ratio), divided by the volume of H₂SO₄). It then requires me to calculate the maximum percentage apparatus uncertainty of the concentration of H₂SO₄. I understand that to calculate the percent uncertainty for my volume calculations it would be uncertainty / volume x 100, adding both for the uncertainty in the volume of NaOH and H₂SO₄, but will I have a greater uncertainty because I had to divide and multiply to find the concentration, and how will I find that uncertainty?

My working so far: The question states that I have to find "the maximum percentage apparatus uncertainty in the final result". I know for sure that I have to then find the percentage uncertainties of my volume calculations:

1. Volume of Sulphuric Acid = 10.0 ± 0.1 mL (cm^3), to find percentage uncertainties:
uncertainty / value x 100 Therefore percentage uncertainty in the volume of sulphuric acid =
0.1 / 10.0 x 100 = 1%

2. Volume of Sodium Hydroxide = 10.73 ± 0.10 mL percentage uncertainty in the volume of sodium hydroxide =
0.10 / 10.73 x 100 = 0.93%

3. I know I must add these values to total 1.93%

4. Now here is where I am confused. Because to find the concentration of H₂SO₄ I had to first, multiply the volume of NaOH (which we know had an uncertainty) by it's concentration to find the moles of NaOH which was then multiplied by 0.5 (mole ratio), which was then divided by the volume of H₂SO₄ (which we know has an uncertainty), how would this procedure affect the total apparatus percentage uncertainty?