500 " mL of " a solution contains 2.65 g of `Na_2CO_3` and 4 g of `NaOH`. 20 " mL of " this solution titrated each time against `(N)/(10)H_2SO_4`. Find out the titre value if (a). Methyl orange is taken as an indicator
(b). Phenolphthalein is taken as indicator.

Explanation

`Na_2CO_4` in `1L=2xx2.65=5.3g`
`NaOH` in `1L=2xx4=8g`
Normality of `Na_2CO_3=(5.3)/(53)=0.1N`
Normality of NaOH`=(8)/(40)=(1)/(5)N`
`N_1V_1(H_2SO_4)=N_2V_2(Na_2CO_3)`
`(N)/(10)xxV_1=20xx0.1N`
`V_1=20cc`
`N_1V_1(H_2SO_4)=N_2V_2(NaOH)`
`(N)/(10)xxV_1=(N)/(5)xx20`
`V_1=(1)/(5)xx20xx10=40cc`
With methyl orange as an indicator:
Full titre value of NaOH`+` Full titre value of `Na_2CO_3`
`=40+20=60cc`
With phenolphthalein as an indicator:
Full titre value of `NaOH+(1)/(2)` titre value of `Na_2CO_3=40+10=50c c`