Question

The calcium contained in a solution of $1.048$ g of a substance being analysed was precipitated with $25$ mL of $H_2{C}_2{O}_4$. The excess of $C_2{O}_4^{2-}$ in one-fourth of filtrate was back titrated with $5$ mL of $0.1025NKMn{O}_4$. To determine the concentration of $H_2{C}_2{O}_4$ solution, it was diluted four folds and the titration of $25$ mL of dilute solution used up $24.1$ mL of same $KMn{O}_4$ solution. Calculate the percentage of $Ca$ in substance.

Answer 1

$Ca+H_2{C}_2{O}_4⟶Ca{C}_2{O}_4\stackrel{KMn{O}_4}{\to }$ colour of $KMn{O}_4$ discharges

The redox changes are as follows:

$(C^{3+}{)}_2⟶2C^{4+}+2e$

$M{n}^{7+}+5e⟶M{n}^{2+}$

$25$ mL of $H_2{C}_2{O}_4$ is used to precipitate $Ca{C}_2{O}_4$.

We need to find the normality of $H_2{C}_2{O}_4$, given that the solution of $H_2{C}_2{O}_4$ is diluted $4$ times, i.e., to $100$ mL.

Meq. of $H_2{C}_2{O}_4$ in $25$ mL dilute solution $=$ Meq. of $KMn{O}_4$ used $=24.1\times 0.1025=2.47025$

$\therefore$ Meq. of $H_2{C}_2{O}_4$ in $100$ mL dilute solution $=\frac{2.47025\times 100}{25}=9.881$

Meq. of $H_2{C}_2{O}_4$ in $25$ mL conc. solution $=9.881$

Meq. of $H_2{C}_2{O}_4$ left after precipitation of $Ca{C}_2{O}_4$ in one fourth filtrate $=$ Meq. of $KMn{O}_4$ used $=5\times 0.1025$

$\therefore$ Total Meq. of $H_2{C}_2{O}_4$ left $=5\times 0.1025\times 4=2.05$

$\therefore$ Meq. of $H_2{C}_2{O}_4$ used for $Ca=9.881-2.05=7.831$
or Meq. of $Ca=7.831$

$\frac{W}{\left(\frac{40}{2}\right)}\times 1000=7.831$

$⟹W_{Ca}=0.1566g$

$\therefore \%$ of $Ca$ in sample $=\frac{0.1566}{1.048}\times 100=14.95\%$

So, the nearest integer value is $15\%$.