Question

Commercial sodium hydroxide weighing 30 g has some sodium chloride in it.
The mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage of sodium chloride in commercial sample of sodium hydroxide? The equation for the reaction is
NaCl + AgNO3 ----> AgCl + NaNO3
[Relative molecular mass of NaCl = 58 ; AgCl = 143]

Answer 1

Dear student,
Let the mass of NaCl in the commercial sample of NaOH be x g.

​The NaCl present in the aqueous solution of the mixture would react with AgNO₃ solution to give white precipitate of AgI.
Molar mass of NaCl = 23 + 35.5 =​58.5g/mol

Molar mass of AgI = 107.8 + 35.5 =143.3 g/mol
NaCl (aq) + AgNO₃ (aq)​-----> AgCl (s) + NaNO₃ (aq)
58.5 g/mol 143.3 g/mol

Thus according to this reaction,
58.5 g NaCl give precipitate = 143.3 g

Therefore x g NaCl will give precipitate = $\frac{143.3}{58.5}\times \mathrm{x}\mathrm{g}$
= 2.449 x g

​Since mass of precipitate obtained = 14.3 g
Thus 2.449 x = 14.3
$\Rightarrow$ x = $\frac{14.3}{2.449}$
$\Rightarrow$ x = 5.839 g

Therefore amount of NaCl present in 30 g of commercial sample of NaOH = 5.839 g
​Percentage of NaCl in sample =$\frac{5.839}{30}\times 100$ = 19.46 %


regards