chem Ksp urgent help !!!

2010-03-25 7:37 am

Solution of 400.0cm^3 of 0.0011 M BaCl2 and 600.0cm^3 of 0.0008M K2SO4 are mixed .
Calculate (I) the concentration of Ba^2+ and SO4^2- ions in the resulting solution.
(II) The amount ( in moles) of the precipitate formed.
What would be the effect of adding approximately 10cm^3 of 1M K2SO4 to the above mixture?
[ Ksp of BaSO4 = 1.1*10^-10 mol ^2 dm^-6 ]

回答 (1)

micatkie

2010-03-25 8:08 am

✔ 最佳答案

(I)
No. of moles of Ba^2+(aq) added = 0.0011 * (400/1000) = 0.00044 mol
No. of moles of SO4^2-(aq) added = 0.0008 * (600/1000) = 0.00048 mol
Volume of the resulting solution = 400 + 600 = 1000 cm^3 = 1 dm^3

Ba^2+(aq) + SO4^2-(aq) → BaSO4(s)
Mole ratio Ba^2+ : SO4^2- = 1 : 1

In the resulting solution after reaction :
No. of moles of SO4^2- left = 0.00048 - 0.00044 = 0.00004 mol
[SO4^2-] = 0.00004/1 = 4 x 10^-5 M ...... (answer)
[Ba^2+] = (1.1 x 10^-10)/(4 x 10^-5) = 2.75 x10^-6 M ...... (answer)

(II)
No. of moles of ppt formed = No. of Ba^2+ ions = 0.00044 mol ...... (answer)
(or 4.4 x 10^-4 mol)

參考: micatkie

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