Question

Question #544e6

Answer 1

0.954g

Explanation:

Here 200mL0.5M HCl contaims
`(200*0.5)/1000=0.1`mole HCl
=0.1gmequivalent HCl

25mLpartly neutralised HCl requires 20.5mL 0.5M NaOH .
forcomlere neutralisation

200mLpartly neutralised HCl requires `20.5*200/25=164mL` 0.5M NaOH for complete neutralisation

164mL0.5M NaOH solution contains
`164*0.5/1000=.082`mole or gm-equivalent NaOH.

Let the mass of `Na_2CO_3`added be x g or `x/53`gm-equivalent
where equivalent mass of`Na_2CO_3=53`
Here total no.of gm-equivalent of base =total no.of gm-equivalent
of acid

`x/53+0.082=0.1`

`x=0.018*53=0.954g`