# What is the solubility of MnS, in grams per liter, in a buffer solution that is 0.105 M CH_3COOH - 0.490 M NaCH_3COO?

## What is the solubility of $M n S$, in grams per liter, in a buffer solution that is 0.105 M $C {H}_{3} C O O H$ - 0.490 M $N a C {H}_{3} C O O$? $M n S$ has a ${K}_{\setminus \textcolor{red}{s p a}}$ of $3 \setminus \cdot {10}^{7}$

Apr 3, 2017

${\text{1.81 g L}}^{- 1}$

#### Explanation:

The idea here is that the solubility of manganese(II) sulfide will increase in an acidic solution because the sulfide anions will act as a base and react with the hydronium cations to form hydrogen sulfide, a weak acid.

In pure water, manganese(II) sulfide will only partially dissociate to produce manganese(II) cations and sulfide anions

${\text{MnS"_ ((s)) rightleftharpoons "Mn"_ ((aq))^(2+) + "S}}_{\left(a q\right)}^{2 -}$

The solubility product constant for this equilibrium reaction is

${K}_{s p} = \left[{\text{Mn"^(2+)] * ["S}}^{2 -}\right]$

Now, it's important to realize that the sulfide anion cannot exist in water. Once it dissociates from the solid, it will react with water to form bisulfide anions, ${\text{HS}}^{-}$.

In other words, the sulfide anions act as a strong base in aqueous solution.

${\text{S"_ ((aq))^(2-) + "H"_ 2"O"_ ((l)) -> "HS"_ ((aq))^(-) + "OH}}_{\left(a q\right)}^{-}$

This means that a more accurate depiction of what happens when manganese(II) sulfide partially dissociates in water would be

${\text{MnS"_ ((s)) + "H"_ 2"O"_ ((l)) rightleftharpoons "Mn"_ ((aq))^(2+) + "HS"_ ((aq))^(-) + "OH}}_{\left(a q\right)}^{-}$

The ${K}_{s p}$ would thus take the form

${K}_{s p} = \left[{\text{Mn"^(2+)] * ["HS"^(-)] * ["OH}}^{-}\right]$

Here comes the cool part. If you have an acidic solution, the excess hydronium cations will neutralize the hydroxide anions produced when the solid partially dissociates.

Moreover, the excess hydronium cations will protonate the bisulfide anions to hydrogen sulfide, $\text{H"_ 2"S}$.

As a result, the equilibrium will shift to the right and more solid will dissolve $\to$ as always, Le Chatelier's Principle governs equilibrium reactions.

This means that in acidic solution, you will have

"MnS"_ ((s)) + 2"H"_ 3"O"_ ((aq))^(+) rightleftharpoons "Mn"_ ((aq))^(2+) + "H"_ 2"S"_ ((aq)) + 2"H"_ 2"O"_ ((l))" "color(purple)("(!)")

This time, the solubility product constant is called the acid solubility product constant, ${K}_{s p a}$, and it takes the form -- remember, pure liquids or solids are not included in the expression of the equilibrium constant!

K_(spa) = (["H"_2"S"] * ["Mn"^(2+)])/(["H"_3"O"^(+)]^2)" "color(darkorange)("(*)")

In your case, you know that the ${K}_{s p a}$ for manganese(II) sulfide is

${K}_{s p a} = 3 \cdot {10}^{7}$

So, know that we've figured out how the balanced chemical equation should look like, focus on determining the concentration of hydronium cations in the buffer.

The $\text{p} {K}_{a}$ of acetic acid is

$\text{p} {K}_{a} = 4.75$

http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Use the Henderson - Hasselbalch equation to find the pH of the buffer

"pH" = "p"K_a + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))

"pH" = 4.75 + log( (0.490 color(red)(cancel(color(black)("M"))))/(0.105color(red)(cancel(color(black)("M")))))

$\text{pH} = 5.42$

As you know, you have

"pH" = - log(["H"_3"O"^(+)])

which implies

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

The concentration of hydronium cations in this solution will be

$\left[{\text{H"_3"O}}^{+}\right] = {10}^{- 5.42} = 3.80 \cdot {10}^{- 6}$

Now, take a look at the balanced chemical equation $\textcolor{p u r p \le}{\text{(!)}}$. Notice that every mole of manganese(II) sulfide that dissociates produces $1$ mole of manganese(II) cations and $1$ mole of hydrogen sulfide.

This means that the two chemical species will have equal concentrations at equilibrium.

The amount of manganese(II) sulfide that dissociates produces manganese(II) cations in a $1 : 1$ mole ratio, which means that if you take $s$ to be this equilibrium concentration, you can say that this value represents the molar solubility of the salt and that equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ becomes

${K}_{s p a} = \frac{s \cdot s}{3.80 \cdot {10}^{- 6}} ^ 2$

which is

$3 \cdot {10}^{7} = {s}^{2} / {\left(3.80 \cdot {10}^{- 6}\right)}^{2}$

All you have to do now is solve for $s$. You will have

${s}^{2} = 3 \cdot {10}^{7} \cdot {\left(3.80 \cdot {10}^{- 6}\right)}^{2}$

which gets you

$s = \sqrt{3 \cdot {10}^{7} \cdot {\left(3.80 \cdot {10}^{- 6}\right)}^{2}} = 0.0208$

You can thus say that in this buffer, the molar solubility of the salt is equal to

$s = {\text{0.0208 mol L}}^{- 1}$

To convert this to grams per liter, use the molar mass of manganese(II) sulfide

color(darkgreen)(ul(color(black)("solubility in g/L"))) = 0.0208 color(white)(.)color(red)(cancel(color(black)("mol")))/"L" * "87.003 g"/(1color(red)(cancel(color(black)("mol MnS")))) = color(darkgreen)(ul(color(black)("1.81 g L"^(-1)))

I'll leave the answer rounded to three sig figs, the number of sig figs you have for the concentrations of acetic acid and acetate anions.