Chem help??
the solubility product of calcium hydroxide Ca(OH)2, is 7.9 x 10^-6 at 25 degrees C. Will a precipitate form when 100 mL of 0.10 mol/L of CaCl2 solution and 50.0 mL of 0.070 mol/L of NaOH solution are combined?
Answer:
moles of Ca2+ from CaCl2 = 0.100 L X 0.10 mol/L = 0.010 moles.
M Ca2+ = 0.010 moles/(0.100 L + 0.050 L) = 0.067 M
moles OH- from NaOH = 0.050 L X 0.070 mol/L = 0.0035 moles.
M OH- = 0.0035 moles/(0.150 L) = 0.0233 M
Ca(OH)2 <---> Ca2+ + 2 OH-, therefore, Qsp = [Ca2+][OH-]^2.
Qsp = (0.067)X(0.0233)^2 = 3.6 X 10^-5. Since Qsp > Ksp, Ca(OH)2 will precipitate from that solution.
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Answer:
moles of Ca2+ from CaCl2 = 0.100 L X 0.10 mol/L = 0.010 moles.
M Ca2+ = 0.010 moles/(0.100 L + 0.050 L) = 0.067 M
moles OH- from NaOH = 0.050 L X 0.070 mol/L = 0.0035 moles.
M OH- = 0.0035 moles/(0.150 L) = 0.0233 M
Ca(OH)2 <---> Ca2+ + 2 OH-, therefore, Qsp = [Ca2+][OH-]^2.
Qsp = (0.067)X(0.0233)^2 = 3.6 X 10^-5. Since Qsp > Ksp, Ca(OH)2 will precipitate from that solution.
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