Question

For 10ml of potassium dichromate solution liberates iodine from potassium iodide solution. When the iodine was titrated with hypo solution $\left( {N/20} \right)$ , the titre value was $15ml$ . Find the concentration of dichromate solution in g per liter.

Answer 1

Hint:
Equivalent weight and the molecular weight of oxidizing or reducing agents will be the same. In iodometric titrations, strong reducing agents that are used are sulphurous acid, hydrogen sulphide and sodium thiosulphate.

Complete step by step answer:
Reaction: $C{r_2}O_7^{2 - } + {I^ - } + 14{H^ + } \to 2C{r^{3 + }} + 3{I_2} + 7{H_2}O$
Hypo Solution also known as $N{a_2}{S_2}{O_3}$ .
When iodine was titrated with the hypo solution $N{a_2}{S_2}{O_3}$ the reaction is given below as follows:
\[{I_2} + 2N{a_2}{S_2}{O_3} \to 2NaI + N{a_2}{S_3}{O_6}\]….1
$1$ mol of potassium dichromate is equivalent to $6$ moles of hypo solution.
The reason for this is because when dichromate reacts with iodine , $6$ moles of iodine is liberated which when further reacts with $6$ moles of hypo solution.
Balancing the equation 2 we get,
\[3[{I_2} + 2N{a_2}{S_2}{O_3}] \to 3[2NaI + N{a_2}{S_3}{O_6}]\].
$n = \dfrac{M}{{MW}}$ but $M = \dfrac{{N \times E \times V}}{{1000}}$
Substituting the value of $M$we will get,
$n = \dfrac{{N \times E \times V}}{{MW \times 1000}}$
Where, $n = $ number of moles, $N = $ normality, $E = $ equivalent mass, $V = $ volume, $MW = $ molecular weight.
Atomic mass of oxygen $ = 16$
Atomic mass of sodium $ = 11$
Atomic mass of sulphur $ = 32$
$E = MW$
Therefore,
$MW$ of hypo solution $N{a_2}{S_2}{O_3} = 2 \times (atomic \, mass \, of \, sodium + atomic \, mass\, of \, sulphur) + 3 \times atomic\, mass\, of \,oxygen$
Substituting the values we get,
$MW$ of hypo solution $N{a_2}{S_2}{O_3} = 2 \times \left( {23 + 32} \right) + 3 \times 16$
$MW$ of hypo solution $N{a_2}{S_2}{O_3} = 110 + 48$
$MW$ of hypo solution $N{a_2}{S_2}{O_3} = 158g$
A.Given data:
$N = \dfrac{1}{{20}}$, $MW = 158$,$E = 158g$, $V = 15ml$.
To find: $n = ?$
Formula to be used: $n = \dfrac{{N \times E \times V}}{{MW \times 1000}}$
Soln:
$n = \dfrac{{N \times E \times V}}{{MW \times 1000}}$
Substituting the values we get,
$n = \dfrac{{1 \times 158 \times 15}}{{20 \times 158 \times 1000}}$
$n = \dfrac{{15}}{{20000}}$
$n = 7.5 \times {10^{ - 4}}mol$
B.Since, $1$mol of potassium dichromate is equivalent to $6$ moles of hypo solution.
Therefore, the number of moles of potassium dichromate $ = \dfrac{1}{6} \times $number of moles of hypo solution.
number of moles of potassium dichromate $ = \dfrac{1}{6} \times 7.5 \times {10^{ - 4}}$
number of moles of potassium dichromate $ = 1.25 \times {10^{ - 4}}mol$
C.$n = \dfrac{M}{{MW}}$
Atomic mass of oxygen $ = 16$
Atomic mass of potassium $ = 39$
Atomic mass of chromium $ = 52$
Molecular weight of ${K_2}C{r_2}{O_7} = 2 \times $ (atomic mass of potassium $ + $ atomic mass of chromium) $ + 7 \times $ atomic mass of oxygen
Molecular weight of ${K_2}C{r_2}{O_7} = 2 \times \left( {39 + 52} \right) + 7 \times 16$
Molecular weight of ${K_2}C{r_2}{O_7} = 182 + 112$
Molecular weight of ${K_2}C{r_2}{O_7} = 294$
Therefore, mass of potassium dichromate $m = MW \times n$
Substituting the values we get,
$m = 1.25 \times {10^{ - 4}} \times 294$
$m = 0.03675g$
Volume of potassium dichromate solution $ = 10ml$
Now, weight of potassium dichromate in gram/liter$ = \dfrac{{0.03675}}{{10}} \times 1000$
weight of potassium dichromate in gram/liter $ = 3.675g/l$
Therefore, the concentration of dichromate solution in g per liter is $3.65g/l$ .

Note:Sodium thiosulphate is a strong reducing agent. It is neutral in nature. Also it is highly used in the treatment in the cyanide poisoning cases. Potassium dichromate is a strong oxidizing agent and is highly used in laboratories for experimental purposes.