標題:
Solubility Product!!!
發問:
Solutions of 400cm3 of 0.0011MBaCl2 and 600cm3 of 0.0008M K2SO4 are mixed. Cal:1.concentration of Ba(2+) and SO4(2-) ions in the resulting solution. 2.the amount of precipitate formed.
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Upon mixing, the initial concentrations for both Ba2+(aq) and SO42-(aq) are diluted as follows: [Ba2+(aq)] = (4/10) x 0.0011 = 0.00044 [SO42-(aq)] = (6/10) x 0.0008 = 0.00048 So using the relations below and solubility product of BaSO4 = 1.08×10-10: ...........................................Ba2+(aq) + SO42-(aq) ? Ba SO4(s) Initial concentration (M):...0.00044........0.00048 Equil. concentration (M):...0.00044 – x...0.00048 – x We have: (0.00044 – x)(0.00048 – x) = 1.08×10-10 x2 - 0.00092x + 2.112×10-7 = 1.08×10-10 x2 - 0.00092x + 2.111×10-7 = 0 x = 4.825 × 10-4 (rejected since concentration of Ba2+ is not as high as it) or 4.375 × 10-4 Therefore, resulting conctrations are: [Ba2+(aq)] = 0.00044 - 4.375 × 10-4 = 2.539 × 10-6 M [SO42-(aq)] = 0.00048 - 4.375 × 10--4 = 4.254 × 10-5 M Moreover, no. of moles of Ba2+/SO42- precipitated = x = 4.375 × 10-4 moles (since the resulting volume is 1 dm3.) Hence mass of precipitate formed = 4.375 × 10-4 × 233.43 = 0.102 grams
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