Question

$30ml$ of a solution containing $9.15g{L}^{-1}$ of an oxalate $K_X{Y}_X(C_2{O}_4{)}_z.n{H}_{2}O$ required for titration $27ml$ of $0.12NNaOH$ and $36ml$ of $0.12NKMn{O}_4$ for oxidation. Find $x,y,z$ and $n$.

Answer 1

$30ml$ solution $9.15g/h$ of ${KnHg\left(C_2{O}_4\right)}_2.n{H}_{2}O$

Using formula

For ${KMnO}_4{N}_1{V}_1=N_2{V}_2$

$36\times 0.12=M\times 2\times Z\times 30ml$

$\left(M\right)\left(Z\right)=\frac{36\times 0.12}{30\times 2}\Rightarrow 0.06\times \frac{6}{5}=\frac{0.36}{5}=0.072⟶\left(1\right)$

Using formula for $NaOH$

$N_1{V}_1=N_2{V}_2$

$0.12\times 27=\left(M\right)\times y\times 30$

$My=0.12\times \frac{9}{10}=0.108⟶\left(2\right)$

We also known from charge balance

$x+y=2Z⟶\left(3\right)$

$\left(1\right)/\left(2\right)$ $\frac{Z}{y}=\frac{2}{3}Z=2/3y⟶\left(5\right)$

From (3) : $x=y/3⟶\left(6\right)$

$M=\frac{9.15}{39x+y+88Z+n18}⟶\left(4\right)$

From $\left(4\right),\left(5\right),\left(6\right),\left(4\right)$ in $\left(1\right)$ & $\left(2\right)$

$\Rightarrow y=3,Z=2,x=1,n=2$