Question

What is the solubility of \(\mathrm{MnS}\), in grams per liter, in a buffer solution that is \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}-0.500 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO} ?\) For \(\mathrm{MnS}, K_{\mathrm{spa}}=3 \times 10^{7}\).

Short answer

The solubility of MnS in the buffer solution should be found from the square root of Kspa when [S^{2-}]=[Mn^{2+}]. Remember to convert this value to grams/liter by multiplying by the molar mass of MnS.

Step-by-step solution

Identify the reaction involved

Manganese(II) sulfide (MnS) dissolves into its ions in water, we can write this dissolution as an equilibrium:\n\(MnS(s) \leftrightarrow Mn^{2+}(aq) + S^{2-}(aq)\)

Write the solubility product expression

The Kspa expression for the dissolution can be written as: \[K_{spa}=[Mn^{2+}][S^{2-}]\] Since the solubility of MnS is minimal, we'll assume that the Mn2+ and S2- concentrations provided by the dissolution are equal, both of which we'll call 's'.

Determine [H+] from the buffer solution

Since this is a buffer solution, the pH is maintained constant by the weak acid/base pair. We can find concentration of [H+] from Henderson-Hasselbalch equation:\[pH = pKa + log\left(\frac{{[A-]}}{{[HA]}}\right)\] Given pKa of acetic acid is 4.74, [A-] as the concentration of CH3COO- and [HA] as the concentration of CH3COOH, we can plug in the values to get [H+].

Find [S^{2-}]

The acetate buffer neutralizes the ions in solution. Hence, for every S2- ion formed there will be H+ ion from the buffer to neutralize it forming HS-. The net equation thus becomes: \[S^{2-}(aq) + 2H^{+}(aq) \leftrightarrow HS^{-}(aq) + H_{2}O(l)\] As [H+] > 's', we can safely say that almost all [S2-] forms HS-. Therefore, [S^{2-}] is equal to the 's' we assumed in Kspa expression.

Determine the solubility of MnS

If [S^{2-}]=[Mn^{2+}]=s and Kspa=3×10^-7, we can substitute these values into the K_{spa} expression: 3×10^-7 = s*s = s^2. Solving for s will give us the solubility of MnS in moles/liter. To convert it to grams/liter, multiply it by the molar mass of MnS.

Key concepts

Buffer Solution

Understanding a buffer solution is key to solving problems involving equilibrium in chemistry. At its core, a buffer solution is composed of a weak acid and its conjugate base, or a weak base and its conjugate acid. This solution resists changes in pH when moderate amounts of acids or bases are added. In the case of the exercise, acetic acid (\(CH_3COOH\)) and sodium acetate (\(NaCH_3COO\)) create a buffer that helps maintain a constant pH.

When a salt like manganese(II) sulfide (\(MnS\)) is dissolved in this buffer, the buffer components will interact with the resultant ions from the salt. This interaction is vital for predicting the solubility of the salt in the buffer, as it influences the ionic equilibrium in the solution. Thus, applying our understanding of buffers is instrumental for correctly calculating the solubility of \(MnS\) in a given buffer.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a formula that relates the pH of a buffer solution to the concentration of the weak acid and its conjugate base. It is mathematically depicted as:
\[pH = pKa + log\left(\frac{{[A^-]}}{{[HA]}}\right)\]
In our exercise, the concentration of acetic acid (\(CH_3COOH\) or HA) and its conjugate base, acetate (\(CH_3COO^-\) or A-), can be plugged into this equation along with the pKa value of acetic acid to calculate the pH. This, in turn, allows for the determination of the hydrogen ion concentration ([H+]), which is crucial for understanding the behavior of [S^{2-}] ions in the buffer solution. This equation is essential for step 3 in the provided step-by-step solution.

Molar Mass

The molar mass represents the mass of a given substance (element or compound) per mole of that substance. It is a bridge between the mass of a sample and the number of moles it contains. The molar mass is typically expressed in units of grams per mole (g/mol) and is calculated by summing the atomic masses of the atoms in a molecule.

In our exercise, after obtaining the solubility in moles per liter ([mol/L]) from the Kspa expression, it is necessary to convert this value into grams per liter ([g/L]) to find the measurable solubility of MnS. This is done by multiplying the solubility value ('s') by the molar mass of MnS. Hence, the molar mass plays a pivotal role in the final step of our solution, allowing us to transition from a theoretical moles-based answer to a practical, weighable amount.

Equilibrium Constant

An equilibrium constant is the ratio of the concentrations of the products to the reactants at equilibrium, each raised to the power of their respective coefficients in the balanced chemical equation. For solubility products, specifically, the equilibrium constant (Kspa) pertains to the dissolution of sparingly soluble salts. It is expressed as:
\[K_{spa} = [M^{n+}]^{m}[X^{m-}]^{n}\]
In the context of dissolution reactions, 'm' and 'n' represent the stoichiometry of the ions produced when the solid salt dissociates in water.

For our exercise involving MnS, once we establish that the concentrations of [Mn^{2+}] and [S^{2-}] are equal, we can set up the Kspa equation with these values squared, since the stoichiometry is 1:1. This K_{spa} value is integral to calculating the solubility of the salt in the solution. The understanding of how equilibrium constants work helps us predict the extent of a reaction under given conditions and is at the heart of the problem-solving process in our step-by-step solution.