Question

A student determines that she used 0.0665 mol of sodium hydroxide (NaOH) to completely titrate 25.00 mL of sulfuric acid solution (H2SO4). What is the molarity of the sulfuric acid?

0.00131 M
0.00262 M
1.31 M
2.62 M

Answer 1

The molarity of the sulfuric acid solution is 1.33 M.

Step-by-step explanation:

To find the molarity of the sulfuric acid solution, we can use the equation:

H2SO4 (aq) + 2NaOH(aq) → Na2SO4 (aq) + 2H2O (l)

From the equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH. The student used 0.0665 mol of NaOH, so the moles of H2SO4 would be half of that, which is 0.03325 mol.

The volume of the sulfuric acid solution used was 25.00 mL, which is equal to 0.02500 L. To find the molarity, we divide the moles of H2SO4 by the volume in liters:

Molarity = moles/volume = 0.03325 mol / 0.02500 L

Molarity = 1.33 M.

Answer 2

For a titration procedure, one must know the reaction going on in the steps involved. For this case, it is the reaction between NaOH and sulfuric acid.

2NaOH + H2SO4 = Na2SO4 + 2H2O

With the reaction and the values given we calculate as follows:

0.0665 mol NaOH ( 1 mol H2SO4 / 2 mol NaOH ) = 0.03325 mol H2SO4

Molarity = 0.03325 / .025 = 1.33 M