A solution of 0.170 M NaOH is used to titrate 24.0 mL of a solution of H2S04:

Question

H2S04(aq) + 2NaOH(ag) → 2H20(1) + Na2SO4 (aq)
If 39.8 mL of the NaOH solution is required to reach the endpoint, what is the molarity of the H2SO4 solution?
Express your answer with the appropriate units.

Answer 1

From the balanced equation, we can see that the mole ratio between NaOH and H2SO4 is 2:1.

We are given the volume of NaOH solution required to reach the endpoint, which is 39.8 mL.

Using the equation M1V1 = M2V2, where M1 is the molarity of the NaOH solution, V1 is the volume of NaOH solution used, M2 is the molarity of the H2SO4 solution, and V2 is the volume of the H2SO4 solution that reacts with the NaOH solution, we can substitute the given values into the equation:

(0.170 M)(39.8 mL) = M2 (24.0 mL)

Simplifying this equation, we get:

6.766 = M2 (24.0 mL)

Dividing both sides of the equation by 24.0 mL, we find:

M2 = 6.766 / 24.0
M2 ≈ 0.2824 M

Therefore, the molarity of the H2SO4 solution is approximately 0.2824 M.