A mixutre solution of KOH and `Na_2CO_3` requires 15 " mL of " `(N)/(20)` HCl when titrated with phenolphthalein as indicator.But the same amoound of the solutions when titrated with methyl orange as indicator requires 25 " mL of " the same acid. Calculate the amount of KOH and `Na_2CO_3` present in the solution.

Explanation

Neutralisation reactions are
`{:(KOH+HCl to KCl +H_(2)O),(Na_(2)CO_(3)+HCl to NaHCO_(3)+NaCl ),(NaHCO_(3)+HCl to NaCl +H_(2)O +CO_(2)):}}`phenolphthalein is used
methyl orange is used .
As discussed in the previous example we have with phenophthalein ,
m.e of 20 mL of N/20 HCl = m.e of KOH + m.e of `Na_(2)CO_(3)`
or m.e or KOH + m.e of `Na_(2)CO_(3)`
or m.e of KoH + m.e of `Na_(2)CO_(3) = 20 xx 1/20 = 1 " "...(1)`
Now , with methyl orange ,
m.e of 30 mL of N/20 hCl
m.e of HOH + m.e of `Na_(2)CO_(3) + " m.e of " Na_(2)CO_(3) = 20 xx 1/20 = 1 " "..(1)`
Now , with methyl orange ,
m.e of 30 mL of N/20 HCl
= m.e of KOH + m.e of `Na_(2)CO_(3) + " m.e of "NaHCO_(3)` produced .
Since m.e of `Na_(2)CO_(3) `= m.e of `Na_(2)CO_(3) ` + m.e of `Na_(2)CO_(3)`
or m.e of KOH + `2 xx ` m.e of `Na_(2)CO_(3) = 1.5 " "...(2)`
Subtracting Eqn. (1) From Eqn. (2) , we get ,
m.e of `Na_(2)CO_(3) = 1.5 - 1 = 0.5`
` :. " equivalent of " Na_(2)CO_(3) = (0.5)/1000`
` :. " wt of " Na_(2)CO_(3) = 106`)
From eqn.s (3) and (1) ,
m.e of KOH ` = 1- 0.5 = 0.5`
Equivalent of KoH ` = (0.5)/1000`
Weight of KOH = `(0.5)/1000 xx 56 = 0.28 g . `( eq. wt of KOH = 56) .