A solution of 0.162 M NaOH is used to titrate 25.0 mL of a solution of H2SO4. If 32.8 mL of the NaOH solution is required to reach the endpoint, what is the molarity of the H2SO4 solution?

Question

Please include the detail such as steps.Thank you

Answer 1

First, always begin with a correctly balanced equation for the reaction taking place.

2NaOH + H₂SO₄ ==> Na₂SO₄ + 2H₂O ... balanced equation (note it takes 2 NaOH for 1 H2SO4)

Next, we find the moles of NaOH that were used:

32.8 ml x 1 L / 1000 ml x 0.162 mol/L = 0.005314 moles NaOH used

Now, using the ration of 2:1 in the balanced equation, we find moles of H2SO4 that were present:

0.005314 mol NaOH x 1 mol H2SO4 / 2 mol NaOH = 0.002657 moles H2SO4 present

Finally, to calculate molarity of H2SO4, we use the definition of molarity of moles/liter to get the answer:

M of H2SO4 = moles/liters and moles = 0.002657 and liters = 25.0 ml x 1 L/1000 ml = 0.025 L

= 0.002657 moles / 0.025 L = 0.106 M (3 sig. figs.)