Excess of `Ag_(2)SO_(4)(s), BaSO_(4)(s)`, and `Ba_(3)(PO_(4))_(2)(s)` are simultaneously in euqilibrium with distilled water. Which of the following is (are) true? Assume no hydrolysis of dissolved ions.

Options

A. `[Ag^(o+)] +2[Ba^(2+)] = 2[SO_(4)^(2-)] +3 [PO_(4)^(3-)]`
B. `2[Ag^(o+)] +4[Ba^(2+)] = 2[SO_(4)^(2-)]+2[PO_(4)^(3-)]`
C. `2[Ag^(o+)]+3[Ba^(2+)]=2[SO_(4)^(2-)] + 2[PO_(4)^(3-)]`
D. `[Ag^(o+)] +[Ba^(2+)] =[SO_(4)^(2-)] +[PO_(4)^(3-)]`

Answer

A

Explanation

`{:(i.Ag_(2)SO_(4),hArr,2Ag^(o+)+,SO_(4)^(2-),),(,,2x,x,),(ii.BaSO_(4),hArr,Ba^(2+)+,SO_(4)^(2-),),(,,y,y,),(iii.Ba_(3)(PO_(4))_(2),hArr,3Ba^(2+)+,2PO_(4)^(3-),),(,,3z,2z,):}`
`:. [Ag^(o+)] = 2x, [SO_(4)^(2-)] = x+y, [Ba^(2+)] = x+3z, [PO_(4)^(3-)] = 2z`
Direct method: By electroneutrality.
`[["Total positive charge"="Total negative charge",],["Total charge"="Charge on the ion"xx"Concentration",]]`
iv. `1 xx [Ag^(o+)] +2[Ba^(2+)] = 2[SO_(4)^(2-)] + 2[PO_(4)^(3-)]`
Alternatively: Substituting the concentration of each ion in (iv).
`2x +2(x +3z) = 2(x+y) +3(2z)`
Dividing by: `x+(x+3z) =(x+y)+(3)/(2)(2z)`
`([Ag^(o+)])/(2) +[Ba^(2+)] = [SO_(4)^(2-)] +(3)/(2) [PO_(4)^(3-)]`
or
`[Ag^(o+)] +2[Ba^(2+)] = 2[SO_(4)^(2-)] +3[PO_(4)^(3-)]`
Hence answer is (a)