Question

40 g of a sample of caustic soda containing NaOH, Na2CO3

and inert impurity is dissolved in

water to prepared 1.0 litre solution. A 25 mL portion of this solution required 23.15 mL 1.022 N HCl

for complete neutralisation. To 25 mL another solution, excess of BaCl2 is added, and resulting

solution required 22.55 mL HCl of same strength to reach the end point. What is the mass percentage

of NaOH and Na2CO3 in the original sample would be.​

Answer 1

Explanation:

Milli-equivalents of Na

2

CO

3

, NaHCO

3

and NaOH in 20 ml solution are 40,20 and 20 respectively.

When phenolphthalein is used from the beginning then,

meq. of HCl = meq. of NaOH + meq. of Na

2

CO

3

Therefore, 1×1×V=(1×20×1)+(1×20×2).

Hence, V=60 ml.

When methyl orange is used from beginning then,

meq. of HCl = (meq. of NaOH + meq. of NaHCO

3

+ 2× meq. of Na

2

CO

3

1×1×V=(1×20×1)+(1×20×1)+(2×1×20×2)

Hence, V=120 ml.

When methyl orange is used after the first end point then,

meq. of HCl =2× meq. of NaHCO

3

1×1×V=(2×1×20×1)

Hence, V=40 ml.