Question

$30$ mL of a solution containing $9.15$ g/L of an oxalate, $K_x{H}_y(C_2{O}_4{)}_z.n{H}_{2}O$ is required for titration $27$ mL of $0.12NNaOH$ and $36$ mL of $0.12NKMn{O}_4$ separately.

Calculate the value of $x$.

Answer 1

Meq. of oxalate salt in $30$ mL $=$ Meq. of $KMn{O}_4$ used $=36\times 0.12$

$\therefore$ Meq. of oxalate salt as reductant in $1$ L $=\frac{36\times 0.12\times 1000}{30}$ $=144$

or $\frac{9.15}{\left(\frac{M}{2z}\right)}\times 1000=144$ ......(i)

$[\because (C_2^{3+}{)}_z⟶(2C^{4+}{)}_z+2ze\therefore E_{salt}=\frac{M}{2z}]$

Now, Meq. of oxalate salt as acid in $30$ mL $=$ Meq. of $NaOH$ used $=27\times 0.12$

$\therefore$ Meq. of oxalate salt as acid $1$ litre $=\frac{27\times 0.12\times 1000}{30}$ $=108$

or $\frac{9.15}{\left(\frac{M}{y}\right)}1000=108$ ...(ii) ($y$ is replaceable H-atom)

By equations (i) and (ii), $\frac{y}{2z}=\frac{3}{4}$ or $4y=6z$ .....(iii)
Also, total cationic charge $=$ total anionic charge
$x+y=2z$ ...(iv)
By equation (iii) and (iv), $x:y:z::1:3:2$
$x$, $y$ and $z$ are in simplest ratio and thus, molecular formula of salt becomes $K{H}_3(C_2{O}_4{)}_2.n{H}_{2}O$.
Also, molar mass of salt $=39+3+176+8n$ $=218+18n$ (v)
Also, by equation (i),

$M=\frac{9.15\times 1000\times 2\times 2}{144}=254.16$ ...(vi)

By equations (v) and (vi), $n=2$
$\therefore$ The formula of oxalate salt is $K{H}_3(C_2{O}_4{)}_2.2H_{2}O$.
The value of $x$ is $1$, $y$ is $3$, $z$ is $2$ and $n$ is $2$.