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:
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
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.
0.0618
To determine the molarity of the original H2SO4 solution, we can use the concept of stoichiometry and the balanced equation provided.
First, let's determine the number of moles of H2SO4 that reacted. We can do this by using the given volumes and concentrations of NaOH and H2SO4.
Molarity (M) = Moles (mol) / Volume (L)
For NaOH:
Molarity of NaOH = 0.100 M
Volume of NaOH = 25.3 mL = 0.0253 L
Moles of NaOH = Molarity of NaOH * Volume of NaOH
= 0.100 mol/L * 0.0253 L
= 0.00253 mol
According to the balanced equation, the stoichiometric ratio between NaOH and H2SO4 is 2:1. This means that 2 moles of NaOH react with 1 mole of H2SO4.
From the equation, we can deduce that 2 moles of NaOH react with 1 mole of H2SO4.
Therefore, the number of moles of H2SO4 that reacted = 0.00253 mol / 2 = 0.001265 mol.
Now, let's calculate the volume of the H2SO4 solution that reacted based on its molarity.
Molarity of final solution = 0.029207 M
Volume of final solution = 18.8 mL = 0.0188 L
Moles of H2SO4 in final solution = Molarity of final solution * Volume of final solution
= 0.029207 mol/L * 0.0188 L
= 0.0005496976 mol
Since the stoichiometric ratio between NaOH and H2SO4 is 2:1, the amount of H2SO4 in the original solution is twice that of the amount reacted.
Moles of H2SO4 in original solution = 2 * Moles of H2SO4 that reacted
= 2 * 0.001265 mol
= 0.00253 mol
Finally, we can calculate the molarity of the original H2SO4 solution by dividing the moles of H2SO4 in the original solution by the volume of the original solution.
Molarity of original H2SO4 solution = Moles of H2SO4 in original solution / Volume of original solution
However, the volume of the original solution is not given in the question. Without the volume information, we cannot determine the exact molarity of the original H2SO4 solution.
So, in conclusion, we need the volume of the original H2SO4 solution to determine its molarity.