Question

A sample of impure tin of mass 0.523 g is dissolved in strong acid to give a solution of Sn2+. The solution is then titrated with a 0.0448 M solution of NO3−, which is reduced to NO(g). The equivalence point is reached upon the addition of 4.05×10−2 L of the NO3− solution.

Please type down the answer on a computer not a piece of paper.

Answer 1

The solution of Sn2+is titrated with a 0.0448 M solution of NO3−, which is reduced to NO(g) as follows:

3 Sn{2+} + 2 NO3{-} + 8 H{+} ---- > 2 NO + 3 Sn{4+} + 4 H2O

number of moles= Moalrity* volume in L

= 0.0448 Mole / L * 4.05×10−2 L

= 0.0018144 mole s NO3−

Calculate the mole of Sn2+ as follows:

0.0018144 mole s NO3−3 mol Sn{2+} / 2 mol NO3{−})

=0.0027216 mol Sn{2+}

Amount in g = number of mole s* molar mass

= 0.0027216 mol Sn{2+}*118.710 g Sn/mol

=0.323 g Sn2+

% of Sn2+ = 0.323 g Sn2+/ sample mass*100

% of Sn2+ = 0.323 g Sn2+/ 0.523*100

=61.76 % Sn