Calculate the $pH$ at which $Mg(OH)_2$ begins to precipitate from a solution containing $0.1$ $M$ $Mg^(2+)$ ions? $K_(sp)$ for $Mg(OH)_2$ = $1.0$ x $10^-11$ .

Risposta:

$"pH" = 9$

Spiegazione:

The idea here is that you need to use magnesium hydroxide's costante di solubilità del prodotto to determine what concentration of hydroxide anions, $"OH"^(-)$ , would cause the solid to precipitate out of solution.

As you know, the dissociation equilibrium for magnesium hydroxide looks like this

$"Mg"("OH")_text(2(s]) rightleftharpoons "Mg"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)$

The solubility product constant, $K_(sp)$ , sarà uguale a

$K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)$

Rearrange to find the concentration of the hydroxide anions

$["OH"^(-)] = sqrt(K_(sp)/(["Mg"^(2+)]))$

Inserisci i tuoi valori per ottenere

$["OH"^(-)] = sqrt((1.0 * 10^(-11))/0.1) = sqrt(1.0 * 10^(-10)) = 10^(-5)"M"$

As you know, you can use the concentration of hydroxide anions to find the solution's pOH

$color(blue)("pOH" = - log(["OH"]^(-)))$

$"pOH" = - log(10^(-5)) = 5$

Finally, use the relationship that exists between pOH and pH a temperatura ambiente

$color(blue)("pOH " + " pH" = 14)$

to find the pH of the solution

$"pH" = 14 - 5 = color(green)(9)$

So, for pH values that are sotto $9$ , the solution will be insaturi. Once the pH of the solution becomes equal to $9$ , la soluzione diventa saturato and the magnesium hydroxide starts to precipitate.

Calculate the pHpH at which Mg(OH)_2Mg(OH)_2 begins to precipitate from a solution containing 0.10.1 MM Mg^(2+)Mg^(2+) ions? K_(sp)K_(sp) for Mg(OH)_2Mg(OH)_2 = 1.01.0 x 10^-1110^-11.

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Calculate the $pH$ at which $Mg(OH)_2$ begins to precipitate from a solution containing $0.1$ $M$ $Mg^(2+)$ ions? $K_(sp)$ for $Mg(OH)_2$ = $1.0$ x $10^-11$ .