Asked by mel-2
Background information: The concentration of various cations in seawater in moles per liter are Na+ (0.46M); Mg2+ (0.056M); Ca2+ (0.01M); Al3+ (4x10^-7M); Fe3+ (2x10^-7M)
Question:
If enough OH- ion is added to precipitate 50% of the Mg2+, what percentage of other the other ions will be precipitated. ( I know the answer is virtually all- but how do I get to that answer?)
Question:
If enough OH- ion is added to precipitate 50% of the Mg2+, what percentage of other the other ions will be precipitated. ( I know the answer is virtually all- but how do I get to that answer?)
Answers
Answered by
DrBob222
Mg = 0.056 M
(Mg^+2)(OH^-)^2 = Ksp
Take a 1 L sample.
You know Mg^+2, divide that by 2, you know Ksp, calculate (OH^-).
Using that OH^-, set up Ksp expressions for the other ions. For example,
(M^+)(OH^-) = Ksp
Plug in Ksp and OH^- and calculate (M^+), then (M^+)/(concn in sea water) = fraction remaining in solution, that subtracted from 1 gives the fraction unpptd. None of the Na^+ will ppt.
(Mg^+2)(OH^-)^2 = Ksp
Take a 1 L sample.
You know Mg^+2, divide that by 2, you know Ksp, calculate (OH^-).
Using that OH^-, set up Ksp expressions for the other ions. For example,
(M^+)(OH^-) = Ksp
Plug in Ksp and OH^- and calculate (M^+), then (M^+)/(concn in sea water) = fraction remaining in solution, that subtracted from 1 gives the fraction unpptd. None of the Na^+ will ppt.
Answered by
mel-2
thanks again
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