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During a titration a student neutralizes 50.0 ml of sulfuric acid with 75.0 ml of 0.25 m sodium hydroxide, as shown below. H2SO4(aq) + 2naoh (aq) → Na2SO4 (aq) + 2h2o (l), What was the poh of the original NaOH solution?
Accepted Answer
To calculate the pOH of the original NaOH solution, we need to first determine the number of moles of NaOH used in the titration. The number of moles of H2SO4 used is 50.0 mL x (1 L/1000 mL) x (0.25 M) = 0.0125 moles. Since 1 mole of H2SO4 reacts with 2 moles of NaOH, the number of moles of NaOH used is:
NaOH moles = 2 x H2SO4 moles
NaOH moles = 2 x 0.0125 moles = 0.025 moles
Now we can calculate the concentration of the original NaOH solution:
NaOH Concentration = moles of NaOH / volume of NaOH (L)
NaOH Concentration = 0.025 moles / (75.0 mL x 1 L/1000 mL) = 0.333 M
Finally, we can calculate the pOH of the original NaOH solution:
pOH = -log[OH-]
Using the Kw expression and the concentration of NaOH, we get:
Kw = [H+][OH-] = 1 x 10^-14
[OH-] = Kw/[H+]
[OH-] = 1 x 10^-14 / 1 x 10^-11
[OH-] = 1 x 10^-3 M
pOH = -log(1 x 10^-3)
pOH = 3.00
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