A solution is prepared by mixing 150.0 mL of 0.200 M Mg2+ and 250.0 mL of 0.100

Question

A solution is prepared by mixing 150.0 mL of 0.200 M Mg2+ and 250.0 mL of 0.100 M F-. A precipitate forms. Calculate the concentrations of Mg2+ and F- at equilibrium. Image below:

Currently having difficulties in solving this question --isn't it fine to just solve for solubility since we are given the Ksp. Or do we have to go through all the stoichiometry? Need explanation thanks!

6. (14 pts) A solution is prepared by mixing 150.0 mL of 0.200 M Mg2 and 250.0 mL of 0.100 M F. A precipitate forms. Calculate the concentrations of Mg2* and F at equilibrium MgE2(s), Mg2+ (aq) + 2F-(aq) Ksp= 6.4 × 10-9

Explanation / Answer

According to the question;

As we know that;

=150mL of 0.2M magnesium ion solution will contain

=(0.2/1000)*150 = 0.03moles of the metal ions.

Similarly,

250mL of 0.1M fluoride ions contain 0.025moles of fluoride ions.

So,

The expression for Ksp is given as Ksp

= (Mg2+)(2F-)^2

Now,

It is given that Ksp= 6.4x10^-9 implying that 6.4x10^-9 = (x)(2x)^2

Where x is the quantity of magnesium fluoride dissolved. Solving for x,

So,

We get 4x^3 = Ksp and x = cube root of (6.4x10^-9)/4 = 1.1696x10^-3 giving the concentrations of magnesium to be 1.1696*10^-3M and that of fluoride to be twice as much, which amounts to 2.3392*10^-3M

So,

The concentrations of mg cations at equilibrium will be 1.1696*10^-3M and that of fluoride will be 2.3392*10^-3M.

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