唔識計chem數呀!!!About percentage purity of magnesium oxide

Question

Magnesium oxide is not very soluble in water, and is difficult to titrate directly. Its purity can be determined by use of "back titration" method. 4.06g of impure magnesium oxide was completely dissolved in 100cm3 of hydrochloric acid, of concentration 2.0 mol dm-3 (in excess). The excess acid required 19.7cm3 of sodium hydroxide (0.20 mol dm-3) for neutralization.

This 2nd titration is called is called a "back-titration", and is used to determine the unreacted acid. Calculate the percentage purity of the magnesium oxide.
(Atomic masses: Mg = 24.3, O = 16.0)

仲有想問一問點為之叫back titration, 唔該哂!!-V-

Answer 1

What is back titration ?

Back titration is an indirect method of titration. In a back titration, the substance being quantitatively determined is completely reacted with an excess of known amount reagent. The quantity of this unreacted reagent is then determined by using a simple titration.

For example, consider the titration in the question. The quantity of magnesium oxide is determined. Magnesium oxide is firstly completely reacted with an excess of known amount of hydrochloric acid. The quantity of unreacted hydrochloric acid is determined by using a simple titration with standard sodium hydroxide solution. The quantity of magnesium oxide is then determined indirectly using the data obtained.

A back titration is used when the substance cannot be quantitatively determined by using simple titration. In this case, magnesium oxide is an insoluble solution, and it is impossible to titrate a solid using a simple titration. Therefore, a back titration is used.

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Calculations :

Firstly, consider the reaction of the excess HCl with NaOH solution :
HCl + NaOH → NaCl + H2O
Mole ratio excess HCl : NaOH = 1 : 1
No. of moles of NaOH used = MV = 0.2 x (19.7/1000) = 0.00394 mol
No. of moles of excess HCl = 0.00394 mol

Then, consider the reaction of MgO with HCl :
MgO + 2HCl → MgCl2­ + H2O
Mole ratio MgO : HCl = 1 : 2
Total no. of moles of HCl = MV = 2 x (100/1000) = 0.2 mol
No. of moles of HCl reacted = 0.2 - 0.00394 = 0.19606 mol
No. of moles of MgO used = 0.196/2 = 0.09803 mol

Molar mass of MgO = 24.3 + 16 = 40.3 g mol-1
Mass of MgO = mol x (molar mass) = 0.09803 x 40.3 = 3.95 g
% by mass of MgO = (3.95/4.06) x 100% = 97.3%