6.5 g mixture of sample containing KOH, NaOH, and `Na_2CO_3` was dissolved in `H_2O` and the volume was made up to 250 mL. 25 " mL of " this solution requires 26.23 " mL of " 0.5 N `H_2SO_4` using methyl orange as indicator, and 19.5 " mL of " same `H_2SO_4` using phenolphathalein as indicator for complete neutralisation. Calculate the percentage of KOH, NaOH, and `Na_2CO_3` in the sample.

Explanation

Let a, b, and c " mL of " `H_2SO_4` is required for neutralisatiobn of KOH, NaOH, and `Na_2CO_3` by using methyl orage as indicator. So
`a+b+c=26.23mL` ..(i)
When phenophthalein is used as indicator,
`a+b+(c)/(2)=19.5` ..(ii)
Solving (i) and (ii), we get
`a+b=12.77mL`
`c=13.46 mL`
m" Eq of "` Na_2CO_3=13.46xx0.5N H_2SO_4`
`=6.73 m" Eq of "Na_2CO_3`
`=6.73xx10^(-3)xx(106)/(2)g]` in 250 mL
`=0357xx(250)/(25)g]` in 250 mL
`=3.57g`
Percentage of `Na_2CO_3=(3.57)/(6.5)xx100=54.92%`
Mass of `(KOH+NaOH)=6.5-3.57=2.93g`
Let x g of KOH and `(2.93-x)g of NaOH`
`((x)/(56)+(2.93-x)/(40))xx1000=12.77xx0.5xx10`
`{:[("Since" a+b=12.77 "in" 25 " mL of " solution),("in" 250 mL, a+b=12.77xx10)]:}`
Solving `x=1.316 g`
`% of KOH=(1.316)/(6.5)xx100=20.24%~
`% of NaOH=100-(54.92+20.24)=24.84%`