How many milliliters of a 0.250 M NaOH solution will be required to completely react with 12.74 mL of 0.250 M H2SO4?

Answer 1

given the balanced equation

H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

we know that for each mole of H₂SO₄, 2 moles of NaOH is needed to react.

given:

Molarity (M) x Volume in Liters (L) = moles (mol)

Molarity (M) x Volume in milliters (mL) = millimoles (mmol)

Millimoles (mmol) / Molarity (M) = Volume in milliters (mL)

moles of H₂SO₄ used:

0.250 (M) x 12.74 (mL) = 3.185 (mmol)

moles of NaOH needed:

3.185 (mmol of H₂SO₄) x (2 moles of NaOH / 1 mole of H₂SO₄) = 6.37 (mmol of NaOH needed)

volume of NaOH needed:

6.37 (mmol) / 0.250 (M) = 25.48 mL

rounding for 3 sigfigs

25.5 mL of 0.250 M NaOH

Answer 2

2NaOH + H2SO4 ==> Na2SO4 + 2H2O ... balanced equation

moles H2SO4 present = 12.74 ml x 1 L/1000 ml x 0.250 mol/L = 0.003185 moles H2SO4

moles NaOH needed to neutralize = 0.003185 mol H2SO4 x 2 mol NaOH/mol H2SO4 = 0.00637 mol NaOH

volume of 0.250 M NaOH needed: (x L)(0.250 mol/L) = 0.00637 moles

x = 0.02548 L = 25.5 mls (to 3 sig figs based on 0.250 M which has 3 sig figs)