Question

Q.3. 0.5 g sample containing MnO2 is treated with HCl, liberating Cl2. The Cl2 is passed into a solution of KI and 30.0 mL of 0.1M Na2S2O3 are required to titrate the liberated iodine. The percentage of MnO2 in sample, atomic weight of Mn being 55, is :

Answer 1

Answer : The purity of $MnO_2$ is, 26.1 %

Explanation :

We have,

$MnO_2+4HCl\rightarrow MnCl_2+Cl_2$

$Cl_2+2KI\rightarrow 2KCl+I_2$

$I2+2Na_2S_2O_3\rightarrow 2NaI+Na_2S_4O_6$

Redox change are :

$2e^-+I_2\rightarrow 2I^-$

$2(S^{2+})_2\rightarrow (S^{5/2+})_4 +2e^-$

$2e^-+Mn^{4+}\rightarrow Mn^{2+}$

From the above reaction we conclude that,

$\text{Meq. of }MnO_2=\text{Meq. of }Cl_2\text{ formed}=\text{Meq. of }I_2\text{ liberated}=\text{Meq. of }Na_2S_2O_3\text{ used}$

Therefore,

$\frac{w}{(M/2)}\times 1000=0.1\times 1\times 30$

The valency factor is, 1

$w=\frac{(0.1\times 1\times 30\times M)}{2000}$

$w=\frac{(0.1\times 1\times 30\times 87)}{2000}$

$w=0.1305g$

The purity of $MnO_2$ = $\frac{0.1305}{0.5}\times 100=26.1\%$

Therefore, the purity of $MnO_2$ is, 26.1 %

Answer 2

Answer:

Explanation:

We have,      

MnO2   +             4HCl       →           MnCl2                +             Cl2

Cl2        +             2KI          →           2KCl                   +             I2

I2         +             2Na2S2O3 →        2NaI                      +             Na2S4O6

Redox change are :

2e           +             I2           →           2I-

2(S2+)2                                 →           (S5/2+)4 +             2e

2e           +             Mn4+      →           Mn2+

The above reaction suggest that,

Meq. of MnO2                   =             Meq. of Cl2formed

=             Meq. of I2 liberated    

=             Meq. of Na2S2O3 used

Therefore,          [w/(M/2)] × 1000             =             0.1 × 1 × 30

[ NNa2S2O3= MNa2S2O3, as, valency factor = 1]

Or,         w            =             (0.1 × 1 × 30 × M)/200

=             (0.1 × 1 × 30 × 87)/200

=             0.1305

Therefore,          Purity of MnO2                  =             (0.1305/0.5) × 100

=             26.1%