Question

A sample containing KBr (M 119&/mol) and KNO3 (M 101 g/mol) was analysed by using the Mohr method, A 1.1250-g sample is dissolved into 200.0 ml solution (Solution A). 20.00-ml. of solution A was titrated with 16.25 ml of 0.OSON AgNOs (M 170) solution. a) Report C of ca solution & according to the tination data b) Calculate C% w/w of KBr in the original sample.

Answer 1

Explanation:

(a)   As we know that relation between normality and volume of two solutions is as follows.

               $N_{1}V_{1} = N_{2}V_{2}$

As the given data is as follows.

             $N_{1}$ = ?,              $N_{2}$ = 0.050 N

              $V_{1}$ = 20 ml,      $V_{2}$ = 16.25 ml

Therefore, putting the given values into the above formula as follows.

                $N_{1}V_{1} = N_{2}V_{2}$

                $N_{1} \times 20 ml = 0.050 N \times 16.25 ml$

                $N_{1}$ = 0.0406 N

Hence, normality of solution A is 0.0406 N.

(b)   As gram equivalent weight of KBr is 119 g/mol.

Formula to calculate normality is as follows.

              Normality = $\frac{weight}{gm equi. wt} \times \frac{1000}{V_{ml}}$

As volume is 200 ml, normality is 0.0406 N, and gram equi. wt is 119 g/mol. Therefore, putting these values into the above formula as follows.

          Normality = $\frac{weight}{gm equi. wt} \times \frac{1000}{V_{ml}}$      

          0.0406 N  = $\frac{weight}{119 g/mol} \times \frac{1000}{200 ml}$

              weight = 0.9668 g

As 0.9668 g of KBr is present in 1.1250 g sample. Hence, weight percentage will be calculated as follows.

                   Weight % of KBr = $\frac{0.9668 g}{1.1250 g}$

                                               = 85.94%

Thus, we can conclude that C% w/w of KBr in the original sample is 85.94%.