0.58 g of `CH_3(CH_2)_(c)COOH` was burnt in excess air and the resulting gases `(CO_2` and `H_2O`) were passed through excess NaOH solution. The resulting solution was divided into two equal parts. One part requires 50 " mL of " 1.0 M HCl for complete neutralisation using phenolphthalein indicator. Another part required 80 " mL of " same HCl for neutralisation using methyl orange as indicator. Calculate the value of n and the amount of excess NaOH solution taken initially.

Explanation

`CH_(3)(CH_(2))_(n)COOH+[(3n+4)/(2)]O_(2)to(n+2)CO_(2)+[(2n+4)/(2)]H_(2)O`
1 " mol of "acid`=(n+2) " mol of "CO_(2)`
`(0.58)/((60+14n)) " mol of "acid=((n+2)xx0.58)/((60+14n))` " mol of "`CO_(2)`
This `CO_(2)` is passed in excess `NaOH` where `CO_(2)` is converted to `Na_(2)CO_(3)` and some NaoH is left
With phenolphthalein:
`(1)/(2) m" Eq of "Na_(2)CO_(3)+m" Eq of "NaOH` left`=m" Eq of "HCl`
`=50xx1xx1`
`=50`
`i.e., (a)/(2)+b=50` ..(i)
With methyl orange
`a+b=80xx1xx1` ..(ii)
By equations (i) and (ii) we get
`b(m" Eq of "NaOH` left)`=20`
`a(m" Eq of "Na_(2)CO_(3) formed)=80-20=60`
Total m" Eq of "`NaOH used=20+60=80`
Weight of NaOH (total)`=80xx10^(-3)xx40=3.2g`
Also `m" Eq of "Na_(2)CO_(3)=m" Eq of "CO_(2)`
`60=((n+2)xx0.58)/((60+14n))xx2`(n-factor)`xx10^(3)`
`thereforen=4`