Question

1.5gm of sample of impure potassium dichromate was dissolved in water and made up to 500mL solution. 25mL of this solution required iodometrically 24mL of a sodium thiosulphate solution. 26mL of this sodium thiosulphate solution required 25mL of $\dfrac{N}{{20}}$ solution of pure potassium dichromate. Find the percentage purity of the impure sample of potassium dichromate.

Answer 1

We can say that we need six moles of sodium thiosulphate in order to titrate one mole of potassium dichromate. The formula to find % purity is as below.
$\% {\text{purity = }}\dfrac{{{\text{Mass of pure }}{{\text{K}}_2}C{r_2}{O_7}}}{{{\text{Total mass of sample}}}}$

Complete step by step solution:
Here, we are given that iodometric titration of potassium dichromate is done in order to estimate the purity of potassium dichromate solution. Iodometric titration of potassium involves the reaction of iodide ions with potassium ions followed by its back titration with sodium thiosulfate solution in order to estimate the number of iodide ions used in the reaction.

- We can write the reaction of potassium dichromate with iodide containing salt like KI as:
${K_2}C{r_2}{O_7} + 6KI + 6{H_2}O \to 2Cr{(OH)_3} + 3{I_2} + 7{H_2}O$
Here, we can see that chromium gets reduced from +6 to +3 oxidation state.

Now, this iodine liberated will be reacted with the solution of sodium thiosulphate in order to measure its quantity. This reaction can be given as:
${I_2} + 2N{a_2}{S_2}{O_3} \to 2NaI + N{a_2}{S_4}{O_6}$
Now, we are given that 26mL of this sodium thiosulphate solution required 25mL of $\dfrac{N}{{20}}$ solution of pure potassium dichromate. Here, we can use the below-given formula to find the concentration of sodium thiosulphate.
${N_1}{V_1} = {N_2}{V_2}$
Here, ${N_1}{\text{ and }}{{\text{V}}_1}$ is normality and volume of sodium thiosulfate solution and ${N_2}{\text{ and }}{{\text{V}}_2}$ is normality and volume of potassium dichromate solution. Putting the available values into this equation, we get
${N_1} \times 26 = \dfrac{1}{{20}} \times 25$
${N_1} = \dfrac{{25}}{{26 \times 20}} = 0.0480N$
Now, we know that the normality of the sodium thiosulphate solution is 0.0480N.
We can write that as 25mL of potassium dichromate solution is treated by 24mL of 0.048N $N{a_2}{S_2}{O_3}$ solution in order to reach end point, if we use 500mL of potassium dichromate solution, then the volume of 0.048N $N{a_2}{S_2}{O_3}$ solution required will be $\dfrac{{500 \times 24}}{{25}} = 480mL$
Number of moles of $N{a_2}{S_2}{O_3}$ in 480mL 0.048N solution = ${\text{Number of moles = }}\dfrac{{N \times V}}{{1000}} = \dfrac{{0.0480 \times 480}}{{1000}} = 0.0230$
We can see from the reactions that six moles of sodium thiosulfate is required to titrate one mole of potassium dichromate solution.
Moles of ${K_2}C{r_2}{O_7}$ = $\dfrac{1}{6}$ moles of $N{a_2}{S_2}{O_7}$ = $\dfrac{1}{6} \times 0.0230 = 3.84 \times {10^{ - 3}}$
Now, we know that molecular weight of ${K_2}C{r_2}{O_7}$ is 294$gmmo{l^ - }$.
So, mass of potassium dichromate in given 1.5gm sample = Moles $\times$ Molecular weight = $3.84 \times {10^{ - 3}} \times 294 = 1.128$
Now, % Purity = $\dfrac{{{\text{Mass of pure }}{{\text{K}}_2}C{r_2}{O_7}}}{{{\text{Total mass of sample}}}} = \dfrac{{1.1289 \times 100}}{{1.5}} = 75.26\%$

Thus, we can conclude that the purity of this sample is 75.26%.

Note: Here, do not forget to calculate the total amount of potassium dichromate present in the given 500mL solution by multiplication as we are given the titration of only 25mL of this solution. Do not get confused between iodimetric and iodometric titration as the later produces iodine upon reaction.