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Mercuric sulfide, HgS, is one of the least soluble salts known, with \(K_{\mathrm{sp}}=1.6 \times 10^{-54}\) at \(25^{\circ} \mathrm{C} .\) Calculate the solubility of HgS in moles per liter and in grams per liter.

Short Answer

Expert verified
The solubility of HgS in water at 25掳C is approximately \( 1.26 \times 10^{-27} \) mol/L or \( 2.93 \times 10^{-25} \) g/L.

Step by step solution

01

Write the Ksp expression

According to the balanced chemical equation, the solubility product constant (Ksp) expression for the dissociation of HgS is given by: Ksp = [Hg^2+][S^2-]
02

Set up the concentration table

Let x mol/L be the molar solubility of HgS in water. Then, the molar concentrations for Hg^2+ and S^2- ions in terms of x are: Hg^2+: x mol/L S^2- : x mol/L
03

Substitute concentrations in Ksp expression

Replace the concentrations of the ions from step 2 in the Ksp expression from step 1: \( K_{sp} = 1.6 \times 10^{-54} = x^2 \)
04

Solve for x

To find the molar solubility (x), solve the equation from step 3 for x: \( x = \sqrt{1.6 \times 10^{-54}} \approx 1.26 \times 10^{-27} \) mol/L
05

Convert moles per liter to grams per liter

To convert the molar solubility (x) into grams per liter, multiply by the molar mass of HgS: Molar mass of HgS = Molar mass of Hg + Molar mass of S = 200.59 g/mol + 32.07 g/mol = 232.66 g/mol Solubility in grams per liter = (1.26 脳 10^{-27} mol/L) 脳 (232.66 g/mol) 鈮 2.93 脳 10^{-25} g/L In conclusion, the solubility of HgS in water at 25掳C is approximately 1.26 脳 10^{-27} mol/L or 2.93 脳 10^{-25} g/L.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Calculations
When it comes to understanding solubility calculations, it's essential to start with the basics. Solubility is a measurement of how much solute can dissolve in a given quantity of solvent at a specific temperature and pressure, resulting in a saturated solution. In the case of sparingly soluble salts like mercuric sulfide (HgS), we often use the solubility product constant (Ksp) to describe their solubility.

Ksp is a special kind of equilibrium constant that applies to the dissolution of solids into ions. For example, the Ksp of HgS (\(1.6 \times 10^{-54}\) at 25掳C) gives us important clues about its solubility. By expressing concentrations in molar units (moles per liter), we can set up a simple equilibrium expression involving the concentrations of the ions produced when HgS dissolves. In the solution process, we establish a direct concentration ratio to Ksp and solve for the molar solubility.

This mathematical relationship allows us to compute the solubility in both moles per liter and grams per liter, changing our approach from abstract constants to practical amounts that can be physically measured in a laboratory setting.
Chemical Equilibrium
Chemical equilibrium is a condition in which the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. It's an essential concept in solubility calculations, especially when dealing with solubility products like Ksp.

In equilibrium, the ions in a saturated solution of a sparingly soluble salt exist in a delicate balance. For HgS, equilibrium is established when solid HgS dissolves to give Hg虏鈦 and S虏鈦 ions and when these ions combine to form solid HgS again.

The beauty of the Ksp lies in its ability to capture this equilibrium state quantitatively. It represents the maximum product of the molar concentrations of the ions that can exist without forming more solid. Understanding this equilibrium allows chemists to manipulate conditions to either favor the dissolution of more solute or the precipitation of excess ions as a solid.
Molar Concentration
Molar concentration, often abbreviated as 'M,' is a measure of the amount of a substance (usually a solute) present in a unit volume of solution and is generally expressed in moles per liter (mol/L). It is crucial in calculating solubility, as it relates the solubility product constant (Ksp) directly to the amounts of each ion present in solution at equilibrium.

In our example with mercuric sulfide, we represent the solubility of HgS using 'x,' which translates to the molar concentrations of Hg虏鈦 and S虏鈦 ions in a saturated solution. By knowing the molar concentration, we can perform various stoichiometric calculations, including converting the molar solubility to grams per liter using the molar mass of the compound.

This conversion is particularly useful since it bridges the gap between abstract chemical concepts and real-world applications, as it allows us to understand and calculate how much substance can be physically dissolved in a solution.

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Most popular questions from this chapter

Under what circumstances can we compare the solubilities of two salts by directly comparing the values of their solubility products?

Write the equilibrium expression for each of the following reactions. a. \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) c. \(\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)\)

Approximately 0.14 g of nickel(II) hydroxide, \(\mathrm{Ni}(\mathrm{OH})_{2}(s),\) dissolves per liter of water at \(20^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

As you know from Chapter \(7,\) most metal carbonate salts are sparingly soluble in water. Below are listed several metal carbonates along with their solubility products, \(K_{\mathrm{sp}} .\) For each salt, write the equation showing the ionization of the salt in water, and calculate the solubility of the salt in mol/L. $$\begin{aligned}&\text { Salt } \quad K_{\mathrm{sp}}\\\&\mathrm{BaCO}_{3} \quad 5.1 \times 10^{-9}\\\&\mathrm{CdCO}_{3} \quad 5.2 \times 10^{-12}\\\&\mathrm{CaCO}_{3} \quad 2.8 \times 10^{-9}\\\&\mathrm{CoCO}_{3} \quad 1.5 \times 10^{-13}\end{aligned}$$

For the process $$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)$$ it is found that the equilibrium concentrations at a particular temperature are \(\left[\mathrm{H}_{2}\right]=1.4 \mathrm{M},\left[\mathrm{CO}_{2}\right]=1.3\) \(M,[\mathrm{CO}]=0.71 \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]=0.66 \mathrm{M} .\) Calculate the equilibrium constant \(K\) for the reaction under these conditions.

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