When 25.3 mL of 0.100 M NaOH was reacted with 18.8 mL of H2SO4 of unknown molarity, the final solution was 0.029207 M in H2SO4. What is the molarity of the original H2SO4 solution? The equation for the reaction is:

Question

2 NaOH + H2SO4 → Na2SO4 + 2 H2O

Answer 1

I like to work in millimols = mmols.
M = mmols/mL. So you ask yourself how much NaOH it would take to neutralize ALL of the H2SO4, not just part of it. How much excess H2SO4 did we have? That's
M H2SO4 = mmols/mL. M = 0.029297. mL in the final solution is 25.3 mL + 18.8 mL = 44.1 mL so mmols in the final solution is 0.029207 x 44.1 = approx 1.3 but that's just an estimate. You must do your own calculations. Convert approx 1.3 mmols H2SO4 to NaOH. That 1.3 x 2 = about 2.6 mmols NaOH. How many mL of 0.1M NaOH is that? M = mmols/mL or mL = mmols/M = about 2.6/0.1 = about 26. So 26 + 25.3 mL or about 51 mL should have neutralized all of the H2SO4. That's a total of 51 x 0.1M = about 5.1 mmols NaOH which is equi8valent to 1/2 that or about 2.6 mmols H2SO4 and that's in 18.8 mL so M H2SO4 = 2.6 mmols/18.8 mL = about 0.14M for the H2SO4.

I never feel right about these problems unless I check them out and you need to do that. What you do is you know that 25.3 mL of 0.1M NaOH was added to 18.8 mL of this 0.14 (not and exact number but use your number) and see if excess H2SO4 is 0.029207 M in the 44.1 mL of the final solution. If it isn't you made a mistake somewhere along the line. I checked it(but I'm not posting it) and it checks out.

Answer 2

0.0618