A 31.5 ml aliquot of H2SO4 was titrated with 0.0134 M NaOH. It took 23.9 ml of base--conctrn of the acid?

Answer 1

Solution:
Always put this principle in your mind:
Milli-equivalents of the Base = Milli-equivalents of the Acid
Translation (using volumes):
(N1)(V1) = (N2)(V2)
Where:
N1 = Normality of Acid
V1 = Volume of Acid used for the reaction
N2 = Normality of the Base
V2 = Volume of the Base (titrant) used for titration
Hence:
(N1)(31.5 ml.) = (0.0134)(23.9 ml.)
N1 = (0.0134)(23.9) / 31.5
N1 = 0.32026 / 31.5
N1 = 0.0102 (Answer) - (Label the bottle: Normality = 0.0102)
Remarks:
(1) Use four decimal places for your subsequent computations.
(2) The concentration of the NaOH solution is given in Molarity (M) but not in Normality (N). When the valence of the metallic component is equal to ONE, the Molarity is equal to the Normality.
(3) If the valence of the metallic component is not equal to ONE, you have to adjust the Normality if the Molarity is given. You do this by knowing the valence of the metallic component.
Trust my solution to this problem. I had been doing titrations for more than two decades!
Email me if you will be faced by this problem.
Good luck.

Answer 2

23.9 mL of 0.0134 mol/L NaOH
= 0.0239 L * 0.0134 mol/L NaOH
= 0.00032026 mol NaOH

NaOH + H2SO4 --> H2O + Na2SO4
Balance this equation and you get:
2NaOH + H2SO4 --> H2O + Na2SO4

Thus, there are 2 mol NaOH for each 1 mol H2SO4
0.00032026 mol NaOH ==> 0.00032026/2 mol H2SO4 = 0.00016013 mol H2SO4

Thus,
0.00016013 mol H2SO4
= 0.0315 L * x mol/L H2SO4

0.00016013 = 0.0315x
x = 0.00016013/0.0315
x = 0.00508

Thus, the concentration is 0.00508 M

Answer 3

Use the molarity of the NaOH and its volume (in Liters) to determine moles of NaOH used.

In this titration, you are reacting 2 moles of NaOH with one mole of H2SO4, so dividing the moles of NaOH by 2 will give you moles of H2SO4.

Divide moles of H2SO4 by volume of H2SO4 (in Liters) to get molarity of H2SO4.