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A mixed solution of potassium hydroxide ...

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 :

A

0.014g

B

0.14g

C

0.028g

D

1.4g

Text Solution

Verified by Experts

The correct Answer is:
A
z`KOH " " + " " Na_(2)CO_(3)`
a m.moles `" " `b m.moles
m.eq. of KOH + m.eq. of `Na_(2)CO_(3)` (v.f.=1)=m.eq. of HCl (in presence of phenolphthalein)
`axx1+bxx1=15xx(1)/(20)`
`:. A+b = 0.75 " " ...(i) ` (in presence of phenolphthalein)
m.eq. of KOH + m.eq. of `Na_(2)CO_(3)` (v.f. =2)=m.eq. of HCl (in presence of methyl orange)
`1xxa+2xxb=25xx(1)/(20)`
`:. a+2xxb=1.25 " " ...(ii)` (in presence of methyl orange)
by solving (i) & (ii), a = 0.25 m.moles.
`:. ` mass of KOH `=(0.25)/(1000)xx56=0.014 g`
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