Question

The concentration of `"OH"^-` ions in a solution left after mixing `100ml` of `0.1M` `"MgCl"_2` and `100ml` of `0.2M` `"NaOH"` (`K_(sp)" of Mg"("OH")_2=1.2*10^-12`)?

Answer 1

`sf([OH^-]=1.34xx10^(-4)color(white)(x)"mol/l")`

Explanation:

If magnesium hydroxide is allowed to come into equilibrium with its ions we get :

`sf(Mg(OH)_((2)(s))rightleftharpoonsMg^(2+)(s)+2OH^(-)(aq)`

`sf(K_(sp)=[Mg_((aq))^(2+)][OH_((aq))^-]^2=1.2xx10^(-12))`

These are equilibrium concentrations.

We now work out the concentrations of the ions we are mixing to give the reaction quotient Q.

`sf(n_(Mg^(2+))=cxxv=0.1xx100/1000=0.01)`

`sf([Mg^(2+)]=n/v=0.01/0.200=0.05color(white)(x)"mol/l")`

`sf(n_(OH^-)=cxxv=0.2xx100/1000=0.02)`

`sf([OH^-]=n/v=0.02/0.200=0.10color(white)(x)"mol/l")`

The reaction quotient Q is given by:

`sf(Q=[Mg_((aq))^(2+)][OH_((aq))^-]^2)`

These are initial concentrations.

`:.` `sf(Q=0.05xx0.10^2=0.0005)`

As you can see this greatly exceeds the value of `sf(K_(sp))`.

If `sf(Q>K)` then the position of equilibrium will shift to the left. This confirms that a precipitate of magnesium hydroxide will form.

Since `sfK` is so small, we can assume that `sf([OH^-])` would be essentially what it would be in a saturated solution in pure water.

Thus, if this is allowed to reach equilibrium with its ions we can say that if `sf([Mg^(2+)]~~x)` then `sf([OH^-]~~2x)`

`:.` `sf(K_(sp)=x.(2x)^2=4x^3=1.2xx10^(-12))`

`sf(x^3=0.3xx10^(-12))`

`sf(x=""^3sqrt(0.3xx10^(-12))=6.7xx10^(-5)color(white)(x)"mol/l")`

This is the concentration of the `sf(Mg^(2+))` ions at equilibrium.

The concentration of `sf(OH^(-))` ions must be twice this.

`:.` `sf([OH^-]=2xx6.7xx10^(-5)=1.34xx10^(-4)color(white)(x)"mol/l")`

Answer 2

Here is a separate approach to this that does it the long way.

I also get `["OH"^(-)] = 1.3_4 xx 10^(-4)`. As it turns out, the fact that `Q_(sp) > K_(sp)`, and that `K_(sp)` is quite small (thus favoring the reactants' side) leads to primarily a saturated solution forming.

This means `["OH"^(-)]` can be determined simply by solving the `K_(sp)` expression in pure water.

(Perhaps if `K_(sp)` wasn't so small, it might matter to do it the long way.)

---

`"MgCl"_2` and `"NaOH"` are strong electrolytes so that, after mixing, the concentration of each relevant species is halved since the volume is doubled:

`["Mg"^(2+)]_i = "0.1 mol"/"L" xx ("0.100 L")/("0.200 L") = "0.05 M"`

`["OH"^(-)]_i = "0.2 mol"/"L" xx "0.100L"/"0.200 L" = "0.1 M"`

With these concentrations, the extent of reaction can be determined:

`Q_(sp) = ["Mg"^(2+)]_i["OH"^(-)]_i^2`

`= ("0.05 M")("0.1 M")^2`

`= 5.0 xx 10^(-4)` `">>"` `K_(sp) = 1.2 xx 10^(-12)`

Since `Q_(sp)` is much bigger than `K_(sp)`, there is too much of the products and the reaction will shift towards the solid reactant to restore equilibrium.

We then treat these concentrations as the initial concentrations in the shift.

`"Mg"("OH")_2(s) rightleftharpoons "Mg"^(2+)(aq) + 2"OH"^(-)(aq)`

`"I"" "-" "" "" "" "" "0.05" "" "" "" "0.1`
`"C"" "-" "" "" "" "-s" "" "" "" "-2s`
`"E"" "-" "" "" "" "0.05-s" "" "" "0.1-2s`

`K_(sp) = 1.2 xx 10^(-12) = (0.05 - s)(0.1 - 2s)^2`

Since `K_(sp)` is so small, we expect `s ~~ "0.05 M"`. Since both terms would go to zero with that assumption, we can't approximate anything here.

However, we could predict that since a saturated solution will form (as the solution would have been overly saturated at this temperature), `["OH"^(-)]` would become what it should be from solving the `K_(sp)` expression in pure water.

Let's see if that's true. The resultant cubic is:

`4s^3 - 0.6s^2 + 0.03s - 0.0005 = 0`

An exact graphical solution (Wolfram Alpha, calculator, etc) would give `s = "0.0499331 M"`, so as a result,

`color(blue)(["OH"^(-)]_(eq)) = "0.1 M" - 2("0.0499331 M")`

`= color(blue)(1.3_4 xx 10^(-4) "M")`

And indeed, we would get the same thing if solving a `K_(sp)` expression (which by definition is for a saturated solution):

`K_(sp) = (s)(2s)^2 = 4s^3`

`=> 2s = ["OH"^(-)]_(eq)`

`= 2(K_(sp)/4)^(1//3) = 1.3_4 xx 10^(-4)` `color(blue)sqrt""`

Answer 3

`["OH"^"-"] = 1.3 × 10^"-4"color(white)(l) "mol/L"`

Explanation:

Our first task is to calculate the stoichiometry of the reaction:

`color(white)(mmmmll)"MgCl"_2 + "2NaOH" → "Mg(OH)"_2 + "2NaCl"`
`"I/mol": color(white)(mll)0.10 color(white)(mmml)0.20color(white)(mmmmll)0`
`"C/mol": color(white)(m)"-0.10"color(white)(mmm)"-0.20"color(white)(mmml)"+0.20"`
`"E/mol": color(white)(mml)0color(white)(mmmmll)0color(white)(mmmmm)0.20`

`"Moles MgCl"_2 = 0.100 color(red)(cancel(color(black)("L"))) × "0.1 mol"/(1 color(red)(cancel(color(black)("L")))) ="0.010 mol"`

`"Moles NaOH" = 0.100 color(red)(cancel(color(black)("L"))) × "0.2 mol"/(1 color(red)(cancel(color(black)("L")))) ="0.020 mol"`

We have stoichiometric amounts, so we end up with a saturated solution of magnesium hydroxide.

`color(white)(mmmmml)"Mg(OH)"_2 → "Mg"^"2+" + "2OH"^"-"`
`"I/mol·L"^"-1": color(white)(mmmmmlmml)0color(white)(mmmll)0`
`"C/mol·L"^"-1": color(white)(mmmmmmm)"+"scolor(white)(mmll)"+2"s`
`"E/mol·L"^"-1": color(white)(mmmmmmmll)scolor(white)(mmmll)2s`

`K_text(sp) = ["Mg"^"2+"]["OH"^"-"]^2 = s(4s)^2 = 4s^3 = 1.2 ×10^"-12"`

`s^3 = (1.2 × 10^"-12")/4 = 3.0 × 10^"-13"`

`s = 6.7 × 10^"-5"color(white)(l) "mol/L"`

`["OH"^"-"] = 2scolor(white)(l) "mol/L" = 2 × 6.7 × 10^"-5" color(white)(l)"mol/L" = 1.3 ×10^"-4" color(white)(l)"mol/L"`