A solution containing 4.2 g of KOH and `Ca(OH)_(2)` is neutralised by an acid. If it consumes 0.1 g equivalents of the acid, calculate the composition of the sample.

Explanation

Let mass of KOH be present in mixture=a g
and Mass of `Ca(OH)_(2)=(4.2-a)g`
Eq. mass of KOH=56, Eq. mass of `Ca(OH)_(2)=(74)/(2)=37`
g equivalent of KOH + g equivalent of `Ca(OH)_(2)`
= g equivalent of the acid
`(a)/(56)+((4.2-a))/(37)=0.1`
or `37a-56a=0.1xx56xx37-4.2xx56`
or 19a=28
`a=(28)/(19)=1.47`
Mass of KOH in the sample =1.47 g
Percentage of KOH = 35
and Percentage of `Ca(OH)_(2)=100-35=65`