Asked by Maddie
If .002 L of .650 M NaOH are added to 1.0 L of 0.9 M CaCl2 what is the value of the reaction quotient and will a precipitate form?
I keep getting 0.0011 because I thought the answer was [Ca+][OH-] but that's not correct, I think I'm missing a final step
I keep getting 0.0011 because I thought the answer was [Ca+][OH-] but that's not correct, I think I'm missing a final step
Answers
Answered by
DrBob222
I can tell you that Ca(OH)2 will not ppt.
Ca(OH)2(s) ==> Ca^+2 + 2OH^-
Qsp = (Ca^+2)(OH^-)^2.
(Ca^+2) = 0.9*(1/1.002) = 0.8982 M which I know is too many places but you can round later.
OH^- = 0.0013*(1.00/1.002) = 0.0013 M
Qsp = (0.8982)(0.0013)^2 = 1.52E-6 which is less than Ksp = 7.9E-6 so no ppt of Ca(OH)2 will occur.
Ca(OH)2(s) ==> Ca^+2 + 2OH^-
Qsp = (Ca^+2)(OH^-)^2.
(Ca^+2) = 0.9*(1/1.002) = 0.8982 M which I know is too many places but you can round later.
OH^- = 0.0013*(1.00/1.002) = 0.0013 M
Qsp = (0.8982)(0.0013)^2 = 1.52E-6 which is less than Ksp = 7.9E-6 so no ppt of Ca(OH)2 will occur.
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