# Chemistry

A solution contains Sr2+ and Ba2+ ions both at a concentration of 1.25 x 10 ^-3 molL-1.
A dilute solution of sulphuric acid is slowly added to this mixture of ions. (Assume any change in volume is insignificant.)
Ks (BaSO4) =1x10^-10 Ks (SrSO4)=3x10^-7

b
(i) calclute the concentration of SO42- ions when SrSO4 begins to precipitate.

(ii) calculate the concentration of Ba2+ when SrSO4 begins to precipitate

thanks(:

 Chic_Bowdrie Posts: 54, Reputation: 31 Junior Member #2 Jul 27, 2012, 01:28 PM
By examining the Ks for each compound, you can tell that BaSO4 is less soluble than SrSO4. So almost all Ba+2 will precipitate out before SrSO4 begins to ppt. Use the Ks for SrSO4 and the initial concentration of Sr+2 to calculate the concentration of SO4-2 for the answer to (i). Then use that concentration of SO4-2 and the Ks for SrSO4 to calculate the concentration of Ba+2 for the answer to (ii). [M+2][SO4-2] = Ks

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