Question

Q4. A solution is prepared by mixing 150.0 mL of 1.00 x 10-M Mg(NO3)2 and
250.0 mL of 1.00 10 M NaF. Calculate the concentrations of Mg.and Fat
equilibrium with solid
| MgF2 (Kg) = 6.4 x 10+).​

Answer 1

Answer:

any options?

Explanation:

Answer 2

Answer:

$[F^-]=0.055M; [Mg^2^+]=2.1*10^-^6M$

Explanation:

Total volume = 150 mL + 250 mL = 400 mL

Initial concentration $[Mg^2^+]=\frac{0.150L*0.01M}{0.400L}=0.00375M$

Initial concentration $[F^-]=\frac{0.250L*0.1M}{0.400L}=0.0625M$

$K_{sp}$ of $6.4*10^-9$ is given for the reaction $MgF_{2}(s)=Mg^2^+(aq)+2F^-(aq)$

And $Q_{sp}=(0.00375)(0.0625)^2=1.465*10^-^5$

So, since $Q_{sp}>K_{sp}$, the reaction shifts towards the reactants ($MgF_2(s)$)

Setting up the ICE table:

$[MgF_2]\,\,\,\,\quad [Mg^2^+]\qquad\quad\,\,[2F^-]\\+x\qquad\quad-x\quad\qquad\quad\,\,\,\,\,-2x\\x\qquad\quad\quad0.00375-x\quad\,\,\,\,0.06250-2x$

Then,

$6.4*10^-^9=(0.00375-x)(0.06250-2x)^2$

which is a cubic function that can be solved by graphing,

giving $x=0.00374788$

$[F^-]=0.0625-2(0.00374788)=0.055M$

$[Mg^2^+]=0.00375-0.00374788=2.1*10^-^6M$

OR

Solved with some extra steps but without a cubic function:

Find the concentrations if the reaction goes to completion based on the stoichiometric ratios:

$Mg^2^++2F^-=MgF_2$

With a concentration of 0.00375M, $Mg^2^+$ is the limiting reactant, and 2(0.00375)=0.0075M of $F^-$ is needed to react with it.

This leaves us with a concentration of:

0M of Mg (all used up)

0.0625-0.0075=0.055M of F (excess)

0.00375M of MgF2

Then, set up an ICE table to find equilibrium in the reverse direction, since in reality, not all the Mg and F will be used up according to the $K_s_p$ value.

$[MgF_2]\,\,\,\,\quad\qquad[Mg^2^+]\qquad\qquad\,\,[2F^-]\\-x\qquad\qquad\quad+x\quad\qquad\qquad\,\,\,\,\,+2x\\0.00375-x\qquad x\qquad\qquad\qquad\quad0.055+2x$

$6.4*10^-^9=(x)(0.055+2x)^2\\\approx(x)(0.055)^2\\\\x\approx\frac{6.4*10^-^9}{(0.055)^2}$

Notice that the 2x can be reasonably discarded in the addition operation due to how small the $K_s_p$ value is (i.e. it is strongly favored towards the reactants), and the concentrations are already sufficiently near equilibrium values (unlike in the previous method shown). If kept, any difference it makes will be below the number of sig figs given.

Solving, $x=2.1*10^-^6$

$[F^-]=0.055$

$[Mg^2^+]=x=2.1*10^-^6$