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10mL of 0.3M Na(2)SO(4) are mixed with 2...

`10mL` of `0.3M Na_(2)SO_(4)` are mixed with `20mL` solution having initially `0.1M Ca^(2+)` and `0.1M Sr^(2+)` in it. Calculate the final `[Ca^(2+)], [Sr^(2+)]` and `[SO_(4)^(2-)]` in solution? Given `K_(sp) SrSO_(4) = 7.6 xx 10^(-7)` and `K_(sp) CaSO_(4) = 2.4 xx 10^(-5)`.

Text Solution

Verified by Experts

`[Ca^(2+)] [SO_(4)^(2-)] = 2.4 xx 10^(-5)`
`[Sr^(2+)][SO_(4)^(2-)] = 7.6 xx 10^(-7)`
`:. ([Sr^(2+)])/([Ca^(2+)]) = 0.03167`
Initial concentration of `SO_(4)^(2-) = (10 xx 0.3)/(30) 0.1M`
Initial concentration of `Sr^(2+) =` Initial concentration of `Ca^(2+) = (0.1 xx 20)/(30) = 0.0667M`
Let final concentration of `Ca^(2+) = X, :. [Sr^(2+)] = 0.03167X`
`:.`Change in concentration of `SO_(4)^(2-)`
= Change in concentration of `Ca^(2+)` and `Sr^(2+)`
`= 0.0667 - X + 0.0667 - 0.03167 X`
`= 0.1334 - 1.03167X`
`:.` Final `[SO_(4)^(2-)] = 0.1 - (0.1334 - 1.03167X)`
[Using `[Ca^(2+] [SO_(4)^(2-)] = K_(sp)` at equilibrium]
`:. X[0.1 - (0.1334 - 1.3167X)] = 2.4 xx 10^(-5)`
`:. 1.03167X^(2) - 0.0334X = 2.4 xx 10^(-5)`
`:. X = 3.3 xx 10^(-2)M or [Ca^(2+)] = 3.3 xx 10^(-2)M`
`:. [Sr^(2+)] = 1.05 xx 10^(-3)M`
`:. [SO_(4)^(2-)] = 7.17 xx 10^(-4)M`
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