Question

0.62g of na2co3. H2o is added to 100ml of 0.1N sol h2so4 the resulting solution will be

Answer 1

Na2CO3.H2O + H2SO4 ---> Na2SO4 + CO2+ 2H2O

given, wt. of Na2CO3 is 0.62g
equivalent wt. = 62g, volume = 100ml

using formula,
weight = NEV/1000
=> 0.62= N*0.62*100/1000
=> N = 0.1N

Since normality of both solutions is equal, resulting solution will be neutral.

Answer 2

Given:

$N{ a }_{ 2 }C{ O }_{ 3 }$ = 0.62 g

${ H }_{ 2 }O$ is added to 100 ml of 0.1 N solution.

To find:

${ H }_{ 2 }{ SO }_{ 4 }$ the resulting solution will be neutral.

Solution:

The chemical reaction between ${ Na }_{ 2 }{ CO }_{ 3 }.{ H }_{ 2 }O\quad and\quad { H }_{ 2 }S{ O }_{ 4 }$ is as follows.

${ Na }_{ 2 }{ CO }_{ 3 }.{ H }_{ 2 }O\quad +\quad { H }_{ 2 }S{ O }_{ 4 }\quad \rightarrow \quad { Na }_{ 2 }S{ O }_{ 4 }\quad +\quad { CO }_{ 2 }\quad +\quad { H }_{ 2 }O$

Let's calculate the equivalent weight of sodium carbonate

Molecular weight of ${ Na }_{ 2 }{ CO }_{ 3 }.{ H }_{ 2 }O = 2(22.9) + 12 + 3(16) + 2(1) + 16 = 123.8 g/mol$

$Equivalent\quad weight\quad of\quad { Na }_{ 2 }{ CO }_{ 3 }.\quad { H }_{ 2 }O\quad =\quad \frac { 123.8 }{ 2 } \quad =\quad 61.9\quad \simeq \quad 62\quad g$

From the given

Weight of ${ Na }_{ 2 }{ CO }_{ 3 }.{ H }_{ 2 }O = 0.62 g$

$Weight\quad =\quad \frac { N\quad \times \quad E\quad \times \quad V }{ 1000 }$

$0.62\quad =\quad \frac { N\quad \times \quad 62\quad \times \quad 100 }{ 1000 }$

$N\quad =\quad \frac { 0.62\quad \times \quad 1000 }{ 62\quad \times \quad 100 } \quad =\quad 0.1$

Hence, Normality of ${ Na }_{ 2 }{ CO }_{ 3 }\quad and\quad { H }_{ 2 }{ SO }_{ 4 }$ is same. Therefore, the solution will be neutral.