1.7225 g of a metal (bivalent) salt `A_(x)(CO_3)_(y)(OH)_(z)` was dissolved in water to make 100 " mL of " solution 50 " mL of " this solution required 10 " mL of " 0.5 M `H_2SO_4` solution for complete neutralisation using phenolphthalein indicator. Another 50 mL solution required 15 " mL of " same acid using methyl orange indicator. Deduce the formula of the salt.

Explanation

Total `+ve` charge`=` Total`-ve` charge
`A_(x)^(2+)(CO_(3)^(2-))y(overset(ɵ)(O)H)_(z):2x=2y+z` .(i)
`H_(2)SO_(4)` reacts with `CO_(3)^(2-)` and `overset(ɵ)(O)H` ions of the salt.
With phenolphthalein:
Half of the salt is neutralised as carbonate is converted to `HCO_(3)^(ɵ)` (bicarbonate ion) and `overset(ɵ)(O)H` ions are completely neutralized.
`therefore{:(m" Eq of "H_(2)SO_(4)),( i n 100 " mL of " solution):}`
`=(1)/(2)m" Eq of salt for "CO_(3)^(2-)+m" Eq of salt of "overset(ɵ)(O)H`
`(10xx1xx100)/(50)=(1)/(2)xx(1.7225xx10^(3))/((M)/(2y))+(1.7225xx10^(3))/((M)/(z))`
`20=(1722.5xxy)/(M)+(1722.5xxz)/(M)` ...(ii)
With methyl orange
Salt is completely neutralised.
`{:[(m" Eq of "H_(2)SO_(4)),(i n 100 " mL of " solu)]:}`
`m" Eq of salt for "CO_(3)^(2-)+m" Eq of salt for "overset(ɵ)(O)H`
`(15xx1xx100)/(50)=(1.7225xx10^(3))/((M)/(2y))+(1.7225xx10^(3))/((M)/(z))`
`30=(1722.5xx2y)/(M)+(1722.5xxz)/(M)` ..(iii)
Subtracting equation (ii) from equation (iii) we get
`10=(1722.5y)/(M)` ..(iv)
By equation (ii) and (iv) we get
`10=(1722.5z)/(M)` ...(v)
By equation (iv) and (v) we get,
`y=z`
Also from equations (i) and (vi), we get
`x=1.5y`
Thus from simplest ratio
`x:y:z::1.5:1:1 or 3:2:2`
Therefore formula of salt is `A_(3)(CO_(3))_(2)(OH)_(2)`