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What mass of \(\mathrm{ZnS}\left(K_{\text {? }}=2.5 \times 10^{-22}\right.\) ) will dissolve in \(300.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} ?\) Ignore the basic properties of \(\mathrm{S}^{2-}\) .

Short Answer

Expert verified
In 300.0 mL of 0.050 M Zn(NO3)鈧 solution and ignoring the basic properties of S虏鈦, the mass of ZnS that will dissolve is approximately \(1.46 \times 10^{-19} \: g\).

Step by step solution

01

Write the balanced equation for the dissolution of ZnS

The dissolution of ZnS in water can be represented by the following balanced equation: \(ZnS (s) \rightleftharpoons Zn^{2+} (aq) + S^{2-} (aq)\)
02

Write the expression for the Ksp of ZnS

The expression for the solubility product constant (Ksp) of ZnS is: \(K_{sp} = [Zn^{2+}] [S^{2-}]\) Given Ksp value is 2.5 脳 10鈦宦猜.
03

Calculate the initial concentration of Zn虏鈦 ions

We know that in 300.0 mL of 0.050 M Zn(NO3)鈧 solution, the initial concentration of Zn虏鈦 ions is 0.050 mol/L.
04

Set up and use ICE table for ZnS

An ICE (Initial, Change, Equilibrium) table can help us keep track of the equilibrium concentrations of the ions. In this case, it will look like this: | | Zn虏鈦 | S虏鈦 | |---------|------|-----| | Initial | 0.050| 0 | | Change | -x | +x | | Equilibrium | 0.050-x | x | Note that we don't know the change in concentration (x) yet, but we can use the given Ksp value to find it.
05

Substitute equilibrium concentrations into the Ksp expression

Substitute the equilibrium concentrations from the ICE table into the Ksp expression: \(K_{sp} = (0.050 - x) (x)\) Plug in the value of Ksp: \(2.5 脳 10^{-22} = (0.050 - x) (x)\)
06

Solve for 'x' to find the solubility of ZnS

This is a quadratic equation, but as the Ksp value is very small, we can consider that x is quite small compared to 0.050. Therefore, the equation can be simplified and solved as follows: \(2.5 脳 10^{-22} = (0.050) (x)\) To solve for 'x' (which corresponds to the concentration of S虏鈦 ions): \(x = \frac{2.5 脳 10^{-22}}{0.050}\) \(x = 5 脳 10^{-21} \: mol/L\)
07

Calculate the mass of ZnS that dissolves

To calculate the mass of ZnS that can dissolve, first convert 'x' (5 x 10鈦宦猜 mol/L) to moles of ZnS for the given volume (300 mL) of the solution: mol ZnS = (5 脳 10鈦宦猜) (0.300 L) mol ZnS = 1.5 脳 10鈦宦猜 mol Now, multiply the number of moles by the molar mass of ZnS: Mass of ZnS = (1.5 脳 10鈦宦猜 mol) (97.45 g/mol) Mass of ZnS = 1.46 脳 10鈦宦光伖 g So, in 300.0 mL of 0.050 M Zn(NO3)鈧 solution and ignoring the basic properties of S虏鈦, 1.46 脳 10鈦宦光伖 g of ZnS will dissolve.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is crucial when studying the solubility of substances. At equilibrium, the rates of the forward and backward reactions are equal, leading to no net change in the concentration of reactants and products. In the context of solubility, this process can be visualized when a solid dissolves and its ions are released into solution at the same rate at which they recombine to form the solid.

Equilibrium does not mean that the reactant and product concentrations are equal, but rather that their ratios remain constant over time. This principle applies to the dissolution of ZnS, where at equilibrium, the product of the concentrations of Zn虏鈦 and S虏鈦 ions equals the solubility product constant, Ksp. The constant is unique for each substance and at a given temperature, providing insight into the compound's solubility.
Solubility Calculations
When performing solubility calculations, the solubility product constant, Ksp, is a pivotal factor. It quantifies the extent to which a compound can dissolve in water. In other words, it is the maximum product of the ion concentrations that can exist without precipitating the solid. To determine the solubility, one would often need to set up an equilibrium equation based on the Ksp to solve for the concentration of ions in solution.

A common misconception is the assumption that a low Ksp always indicates a low solubility. While this is frequently the case, it's also crucial to consider the stoichiometry of the dissolution reaction as it affects the moles of each ion produced in solution. For example, a dissolution equation producing multiple ions could have a higher solubility despite a low solubility product.
ICE Table Method
The ICE table method鈥攕tanding for Initial, Change, Equilibrium鈥攁llows students to systematically calculate the changes in concentration as a reaction proceeds towards equilibrium. It's an invaluable tool for solubility calculations, as seen with the thorough textbook solutions for ZnS.

For ZnS, setting up an ICE table starts with initial ion concentrations, including those from other sources like Zn(NO3)鈧 in our exercise. Then, we define the change in concentrations using a variable, such as 'x', before calculating the equilibrium concentrations. This approach simplifies the substitution into the Ksp expression to compute the solubility. Although sometimes approximations are used when 'x' is small, making sure the approach is valid based on the initial concentrations contributes to a more accurate calculation.

To enhance students' grasp of this concept, educators should emphasize the logical progression through the ICE table and the justification for simplifications in the calculations.

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Most popular questions from this chapter

The copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times 10^{-6}\). Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}\) : $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 \mathrm{M} \mathrm{NaCl}\).

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{~mol} / \mathrm{L}\). b. The solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\).

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?

\(\mathrm{Mg}(\mathrm{OH})_{2}\) is the main ingredient in the antacid TUMS and has a \(K_{\text {sp }}\) value of \(8.9 \times 10^{-12}\). If a \(10.0-\mathrm{g}\) sample of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is placed in \(500.0 \mathrm{~mL}\) of solution, calculate the moles of \(\mathrm{OH}^{-}\) ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is small, not a lot of solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acids.

Describe how you could separate the ions in each of the following groups by selective precipitation. a. \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\) c. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\) b. \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\)

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