Question #544e6

1 Answer
May 2, 2016

Answer:

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#