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

Answer

A

Explanation

`underset(a Mili eq.)(KOH)+underset(b Mili eq.)(Na_2CO_3)`
`a+b/2=15xx1/20`
2a+b=1.5 ..(i) (in presence of phenolphthalein)
`a+b=25xx1/20=1.25`…(ii)(in presence of Methyl orange)
by solving (i) & (ii) a = 0.25 Mili eq.
mass of KOH=`0.25/1000xx56=0.014` gm