Calculation on titration

To do such titration questions, you should be very careful to some keywords. For example, "completely neutralized" means both reactants react with each other completely without any present of excess substances (unreacted).

First, you need to write down the equations of this reaction.

2NaOH (aq) + H2SO4 (aq) --} Na2SO4 (aq) + 2H2O (l)

Second, you are given the volume and molarity (concentration) of H2SO4.

Therefore, you can find no. of mole of H2SO4.

20/1000 * 1 = 0.02 mole

By means of complete reaction, ratio of no. of moles of H2SO4 to no. of mole of Na2SO4 is 1:1

Hence, no. of mole of Na2SO4 is 0.02 * 1 = 0.02 mole

* Since Na2SO4 is soluble in water, we can simply take sodium sulphate and water formed into one substance. Therefore, the volume of resultant solution is equal to volume of NaOH and H2SO4 used.

But how to calculate volume of NaOH?

Easy. Ratio of no. of moles of H2SO4 to no. of moles of NaOH is 1:2 while molarity of NaOH is 2M.

Therefore, no. of mole of NaOH = 0.02 * 2 = 0.04 mole

Hence, volume of NaOH = 0.04 / 2 = 0.02 dm^3

Now, we have 0.02dm^3 of NaOH, 0.02dm^3 of H2SO4 and 0.02 mole of Na2SO4

Concentration of Na2SO4,

no. of mole of Na2SO4 / volume of Na2SO4 (i.e. volume of H2SO4 + NaOH)

= 0.02 / (0.02+0.04)

= 0.33M (corr. to 2 d.p.)

The answer is B.




Hope my explanations help.

2010-03-11 13:38:14 補充：
Sorry that I have made some mistakes,

Concentration of Na2SO4,

no. of mole of Na2SO4 / volume of Na2SO4 (i.e. volume of H2SO4 + NaOH)

= 0.02 / (0.02+0.02)

= 0.50 M (corr. to 2 d.p.)

The answer should be C.

Sorry for that~

2NaOH + H2SO4-->Na2SO4 + 2H2O
no of mole of H2SO4 = (20/1000)(1) = 0.02mol
since it is a complete reaction
thus no of mole of NaOH = 0.02 x 2 = 0.04mol
vol of NaOH = 0.04 / 2 = 0.02dm^3 = 20cm^3

no of mole of Na2SO4 = 0.02mol
[Na2SO4] = 0.02 / (20+20/1000)
= 0.5M

therefore the ans is C

hope i can help u