0.8 Back titration  (Page 2/2)

Problem : 2.75 gram of a sample of dolomite containing $CaC{O}_3$ and $MgC{O}_3$ is dissolved in 80 ml 1N HCl solution. The solution is then diluted to 250 ml. 25 ml of this solution requires 20 ml of 0.1 N NaOH solution for complete neutralization. Calculate percentage composition of the sample.

Solution : We shall first determine the amount of HCl required to react with the sample by using back titration data. Here, we make use the fact that dilution does not change meq of a solution.

$$\Rightarrow \text{meq of 25 ml of diluted excess HCl solution}=\text{meq of NaOH solution}=0.1X20=2$$

$$\Rightarrow \text{meq of 250 ml of diluted excess HCl solution}=20$$

$$\Rightarrow \text{total meq of HCl}=NV=80X1=80$$

$$\text{meq of HCl used to react with dolomite}=80-20=60$$

Mass of HCl used in reaction with dolomite is calculated using expression of meq :

$$meq=\frac{g}{E}X1000=\frac{xg}{M_O}X1000$$

$$\Rightarrow g=\frac{meqX{M}_O}{1000x}=\frac{60X36.5}{1000X1}=2.19gm$$

In order to find the composition, we apply mole concept. The reactions involved are :

$$CaC{O}_3+2HCl\to CaC{l}_2+C{O}_2+H_{2}O$$

$$MgC{O}_3+2HCl\to MgC{l}_2+C{O}_2+H_{2}O$$

Let mass of $CaC{O}_3$ in the sample is x gm. Then mass of $MgC{O}_3$ is 2.75-x gm.

$$\frac{x}{M_{CaC{O}_3}}\text{moles of}CaC{O}_3=\frac{x}{100}\text{moles of}CaC{O}_3=\frac{2x}{100}\text{moles of HCl}$$

$$\Rightarrow \text{mass of HCl required for}CaC{O}_3=\frac{2xX{M}_{HCl}}{100}=\frac{2xX36.5}{100}=\frac{73x}{100}=0.73xgm$$

Similarly for $MgC{O}_3$ ,

$$\Rightarrow \text{mass of HCl required for}MgC{O}_3=\frac{2\left(2.75-x\right)M_{HCl}}{84}=\frac{2X\left(2.75-x\right)36.5}{84}=\frac{73\left(2.75-x\right)}{84}=2.39-0.87xgm$$

$$\Rightarrow \text{total HCl required}=0.73x+2.39-0.87x=2.19$$

$$\Rightarrow 0.14x=0.2$$

$$x=1.43gm$$

$$\Rightarrow \text{\% mass of}CaC{O}_3=\frac{1.43}{2.75}X100=52$$

$$\Rightarrow \text{\% mass of}MgC{O}_3=100-52=48$$

Problem : A sample of $MnS{O}_{\mathrm{4,}}4H_{2}O$ (molecular wt : 223) is heated in air. The residue is ${\mathrm{Mn}}_3{O}_4$ , whose valency and molecular weight are 2 and 229 respectively. The residue is dissolved in 100 ml of 0.1 N $FeS{O}_4$ solution containing dilute sulphuric acid. The resulting solution reacts completely with 50 ml of $KMn{O}_4$ solution. 25 ml of the $KMn{O}_4$ solution requires 30 ml of 0.1N $FeS{O}_4$ solution for complete oxidation. Calculate the weight of $MnS{O}_{\mathrm{4,}}4H_{2}O$ in the given sample.

Solution : The excess of $FeS{O}_4$ completely reacts with 50 ml of $KMn{O}_4$ solution of unknown concentration. We need to know this concentration. From the data of complete oxidation, however, we can know concentration. Let 1 and 2 subscripts denote $KMn{O}_4$ and $FeS{O}_4$ , then

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

$$\Rightarrow N_{1}X25=0.1X30$$

$$\Rightarrow N_1=\frac{3}{25}=0.12N$$

Now, considering back titration,

$$\Rightarrow \text{total meq of}FeS{O}_4\text{solution}=0.1X100=10$$

$$\Rightarrow \text{meq of excess}FeS{O}_4\text{solution}=0.12X50=6$$

$$\Rightarrow \text{meq of}FeS{O}_4\text{solution used for residue}=10-6=4$$

$$\Rightarrow \text{meq of}M{n}_2{O}_3\text{in the residue}=4$$

Mass of ${\mathrm{Mn}}_3{O}_4$ is determined using meq expression :

$$meq=\frac{g}{E}X1000=\frac{xg}{M_O}X1000$$

$$\Rightarrow g=\frac{meqX{M}_O}{1000x}=\frac{4X229}{1000X2}=0.458gm$$

The decomposition of $MnS{O}_{\mathrm{4,}}4H_{2}O$ takes place as :

$$3MnS{O}_{\mathrm{4,}}10H_{2}O\to M{n}_3{O}_4+3S{O}_2+10H_{2}O$$

Applying mole concept, the amount of $MnS{O}_{\mathrm{4,}}4H_{2}O$ is :

$$\Rightarrow \text{Mass of}MnS{O}_{\mathrm{4,}}4H_{2}O=\frac{3X223}{229}X0.458=1.338gm$$