A mixed solution of potassium hydroxide and sodium carbonate required 15 mL of an N/20 HCl solution when titrated with phenolphthalein as an indicator. But the same amount of the solution, when titrated with methyl orange as an indicator , required 25 mL of the same acid. The amount of KOH present in the solution is :

Options

A. `0.014g`
B. `0.14g`
C. `0.028g`
D. `1.4g`

Answer

A

Explanation

Let milli moles of `KOH = x` and `Na_(2)CO_(3) = y`
`x +y = 15 xx (1)/(20)` (with phenolphthalein)
`2x +2y = 1.5` .........(1)
`x 2+2y = 25 xx (1)/(20)` (with methyl orange)
`2x +4y = 2.5` ......(2)
`2y = 1 rArr y = 0.5` milli mole
`x = 0.25` milli moles
weight of `KOH = 0.25 xx 56 xx 10^(-3) = 0.014g`