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Chapter 15: Problem 93

Consider the reaction \(\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{4} \mathrm{IO}_{6}{ }^{-}(a q)\); \(K_{c}=3.5 \times 10^{-2}\). If you start with \(25.0 \mathrm{~mL}\) of a \(0.905 \mathrm{M}\) solution of \(\mathrm{NaIO}_{4}\), and then dilute it with water to \(500.0 \mathrm{~mL}\), what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}{ }^{-}\)at equilibrium?

Short Answer

Expert verified
The equilibrium concentration of H鈧処O鈧嗏伝 is approximately \(1.52 \times 10^{-3}\) M.

Step by step solution

01

Calculate the initial concentration of IO鈧刕- after dilution.

To find the initial concentration of IO鈧刕- in the final solution after dilution, we can use the equation: C1V1 = C2V2 Where: C1 = initial concentration of IO鈧刕- V1 = initial volume of NaIO鈧� solution C2 = final concentration of IO鈧刕- V2 = final volume of NaIO鈧� solution Plug in the given values to find C2: (0.905 M)(25.0 mL) = C2(500.0 mL) C2 = (0.905 M * 25.0 mL) / 500.0 mL C2 = 0.04525 M The concentration of IO鈧刕- after dilution is 0.04525 M.
02

Write down the balanced chemical equation and set up the ICE table.

Our balanced chemical equation is: IO鈧刕- (aq) + 2 H鈧侽 (l) 鈬� H鈧処O鈧哵- (aq) Set up the ICE table: | Species | Initial (M) | Change (M) | Equilibrium (M) | |---------|-------------|------------|-----------------| | IO鈧刕- | 0.04525 | -x | 0.04525 - x | | H鈧侽 | | | | | H鈧処O鈧哵- | 0 | +x | x | Here, x represents the change in concentration of IO鈧刕- and H鈧処O鈧哵-. Notice that we have not included H鈧侽 in the ICE table, as its concentration in the liquid form does not affect the reaction's equilibrium constant.
03

Write the expression for Kc and plug in the equilibrium concentrations.

The expression for Kc is as follows: \[K_c = \frac{[H鈧処O鈧哵-]}{[IO鈧刕-]}\] Given the equilibrium constant Kc = 3.5 脳 10^(-2): \[3.5 \times 10^{-2} = \frac{x}{0.04525 - x}\]
04

Solve for x, the equilibrium concentration of H鈧処O鈧哵-.

We need to solve the equation for x: \[3.5 \times 10^{-2} = \frac{x}{0.04525 - x}\] Let us first multiply both sides by (0.04525 - x) to simplify the equation: \(x = 3.5 \times 10^{-2} (0.04525 - x)\) Now, let's expand and rearrange the equation: \(x = (3.5 \times 10^{-2})(0.04525) - (3.5 \times 10^{-2})x\) Combine x terms: \(x + (3.5 \times 10^{-2})x = (3.5 \times 10^{-2})(0.04525)\) Factor out x and solve: \(x(1 + 3.5 \times 10^{-2}) = (3.5 \times 10^{-2})(0.04525)\) Divide by (1 + 3.5 脳 10^(-2)): \(x = \frac{(3.5 \times 10^{-2})(0.04525)}{1 + 3.5 \times 10^{-2}}\) Calculate the value of x: \(x \approx 1.52 \times 10^{-3}\) The equilibrium concentration of H鈧処O鈧哵- is approximately 1.52 脳 10^(-3) M.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant (\(K_c\)) is a ratio that provides a quantitative measure of the position of equilibrium for a reversible chemical reaction at a given temperature. For the reaction where reactants A and B form products C and D, the equilibrium constant expression is given as \( K_c = \frac{[C][D]}{[A][B]} \) where the concentrations are those at equilibrium, and the brackets denote molarity. The larger the value of the equilibrium constant, the more product there is relative to reactant at equilibrium. In our textbook exercise, the value of \( K_c \) for the reaction \( IO_4^- + 2 H_2O \rightleftharpoons H_4IO_6^- \) is \( 3.5 \times 10^{-2} \), which indicates that at equilibrium, the concentration of reactants is higher than that of the products.
ICE Table
The ICE table, which stands for Initial, Change, and Equilibrium, is a tool used to organize data and find the equilibrium concentrations of reactants and products in a chemical reaction. It's a systematic way to apply stoichiometry to reactions at equilibrium. In our problem, we created an ICE table for the reaction \( IO_4^- + 2 H_2O \rightleftharpoons H_4IO_6^- \), identifying the initial concentrations, the changes during the reaction (expressed as '-x' for reactants and '+x' for products), and the equilibrium concentrations. Since the solvent water is a pure liquid its concentration is not included in the ICE table as it remains essentially constant during the reaction.
Dilution Calculations
Dilution calculations are used to determine the new concentration of a solution after it has been diluted with additional solvent. The core principle used is that the amount of solute remains the same before and after dilution. The equation \( C_1V_1 = C_2V_2 \) encapsulates this idea, with \( C_1 \) and \( V_1 \) representing the initial concentration and volume, and \( C_2 \) and \( V_2 \) the final concentration and volume, respectively. In the context of our exercise, we used this equation to find the concentration of \( IO_4^- \) after diluting \( NaIO_4 \) solution from 25.0 mL to 500.0 mL.
Chemical Reaction Equilibrium
Chemical reaction equilibrium occurs when 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. This dynamic state can be described by the equilibrium constant (\(K_c\)), which remains constant at a given temperature. An important point to note, as demonstrated by the textbook exercise, is that pure solids and liquids do not appear in the expression for the equilibrium constant, as their concentrations are unchanged.
Solving for Equilibrium
Solving for equilibrium involves finding the concentrations of all species at equilibrium. From the ICE table, we can express these concentrations in terms of a variable \( x \). Then, we use the equilibrium constant expression to write an equation in \( x \) and solve for it. As seen in the given problem, after setting up the expression for \( K_c \) and inserting our \( x \) values from the ICE table, we end up with a mathematical equation. Solving this equation gives us the value of \( x \) that corresponds to the concentration of \( H_4IO_6^- \) at equilibrium. It's sometimes necessary to solve a quadratic equation, but in our problem, we could simplify the expression to isolate \( x \) and find the answer.

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Most popular questions from this chapter

Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{3}(s)\). (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) All the compounds in this reaction are soluble in water. Rewrite the equilibriumconstant expression in terms of molarities for the aqueous reaction

A chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate molecules bind to a protein involved in cancer. The drug molecules bind the protein in a \(1: 1\) ratio to form a drugprotein complex. The protein concentration in aqueous solution at \(25^{\circ} \mathrm{C}\) is \(1.50 \times 10^{-6} \mathrm{M}\). Drug \(\mathrm{A}\) is introduced into the protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\). Drug \(B\) is introduced into a separate, identical protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\). At equilibrium, the drug A-protein solution has an A-protein complex concentration of \(1.00 \times 10^{-6} \mathrm{M}\), and the drug \(\mathrm{B}\) solution has a B-protein complex concentration of \(1.40 \times 10^{-6} \mathrm{M}\). Calculate the \(K_{c}\) value for the \(A\)-protein binding reaction and for the \(\mathrm{B}\) protein binding reaction. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?

Consider the reaction \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\). Assume that both the forward reaction and the reverse reaction are elementary processes and that the value of the equilibrium constant is very large. (a) Which species predominate at equilibrium, reactants or products? (b) Which reaction has the larger rate constant, the forward or the reverse?

Write the expression for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g)\) (b) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) (c) \(\mathrm{Ni}(\mathrm{CO})_{4}(g) \rightleftharpoons \mathrm{Ni}(s)+4 \mathrm{CO}(g)\) (d) \(\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}(a q)\) (e) \(2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(a q) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{Zn}(s)\) (f) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (g) \(2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+2 \mathrm{OH}(a q)\)

At \(1285^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is \(K_{c}=1.04 \times 10^{-3}\). A \(0.200-\mathrm{L}\) vessel containing an equilibrium mixture of the gases has \(0.245 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel?
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