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Chapter 12: Problem 89

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is \(\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \quad K=5.6 \times 10^{-11}\) If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into \(1.00 \mathrm{~L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}{ }^{2-}\) ?

Short Answer

Expert verified
The equilibrium concentration of \(\mathrm{CO}_{3}^{2-}\) will be approximately \(3.003 \times 10^{-6}\) M.

Step by step solution

01

Write down the given information

From the exercise, we are given: - The equilibrium reaction: \(\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)\) - The equilibrium constant: \(K = 5.6 \times 10^{-11}\) - Initial concentration of \(\mathrm{HCO}_{3}^{-}\): 0.16 mol/L
02

Set up an ICE table

An ICE table helps us track the changes in concentrations of each species involved in the reaction. ICE stands for Initial, Change, and Equilibrium. We'll fill in the table with the given information: | | \(\mathrm{HCO}_{3}^{-}\) | \(\mathrm{H}^{+}\) | \(\mathrm{CO}_{3}^{2-}\) | |------- |--------------|--------------|--------------| | Initial (M) | 0.16 | 0 | 0 | | Change (M) | -x | +x | +x | | Equilibrium (M)| 0.16 - x | x | x |
03

Write the equilibrium expression

Using the given equilibrium constant, we can write the equilibrium expression: \(K = \frac{[\mathrm{H}^{+}][\mathrm{CO}_{3}^{2-}]}{[\mathrm{HCO}_{3}^{-}]}\) Now, substitute the equilibrium concentrations from the ICE table into the expression: \(5.6 \times 10^{-11} = \frac{x \cdot x}{0.16 - x}\)
04

Solve for x (the equilibrium concentration of \(\mathrm{CO}_{3}^{2-}\))

To solve for x, we can simplify the expression by multiplying both sides by \((0.16 - x)\): \(x^2 = (5.6 \times 10^{-11}) (0.16 - x)\) Since K value is very small, we know that the reaction barely proceeds forward and the amount of x should be extremely small as it forms very little products. So, we can make the approximation that \((0.16 - x) \approx 0.16\): \(x^2 \approx (5.6 \times 10^{-11})(0.16)\) Now, solve for x: \(x \approx \sqrt{(5.6 \times 10^{-11})(0.16)}\) \(x \approx 3.003 \times 10^{-6}\)
05

Determine the equilibrium concentration of \(\mathrm{CO}_{3}^{2-}\)

Now that we have the value of x, we can find the equilibrium concentration of \(\mathrm{CO}_{3}^{2-}\), which, according to the ICE table, is equal to x: \([\mathrm{CO}_{3}^{2-}]_{eq} \approx 3.003 \times 10^{-6} \mathrm{M}\) At equilibrium, the concentration of \(\mathrm{CO}_{3}^{2-}\) will be approximately \(3.003 \times 10^{-6}\) M.

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Most popular questions from this chapter

Calculate a value for the equilibrium constant for the reaction $$\mathbf{O}_{2}(g)+\mathbf{O}(g) \rightleftharpoons \mathbf{O}_{3}(g)$$.given $$\begin{aligned}& \mathrm{NO}_{2}(g) \stackrel{h v}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) & & K=6.8 \times 10^{-49} \\\\\mathrm{O}_{3}(g)+\mathrm{NO}(g) & \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & & K=5.8 \times 10^{-34}\end{aligned}$$.(Hint: When reactions are added together, the equilibrium expressions are multiplied.) (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

At a particular temperature, \(K=1.00 \times 10^{2}\) for the reaction$$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$$.In an experiment, 1.00 mole of \(\mathrm{H}_{2}, 1.00\) mole of \(\mathrm{I}_{2}\), and 1.00 mole of HI are introduced into a 1.00 -L container. Calculate the concentrations of all species when equilibrium is reached.

Write expressions for \(K\) and \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\). b. \(2 \mathrm{NBr}_{3}(s) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\). c. \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\). d. \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\).

At \(25^{\circ} \mathrm{C},\) gaseous \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\) to the extent that \(12.5 \%\) of the original \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is 0.900 atm. Calculate the value of \(K_{\mathrm{p}}\) for this system.

Consider the decomposition of the compound \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) as follows:$$\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g)$$.When a 5.63 -g sample of pure \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g)\) was sealed into an otherwise empty 2.50 -L flask and heated to \(200 .^{\circ} \mathrm{C},\) the pressure in the flask gradually rose to 1.63 atm and remained at that value. Calculate \(K\) for this reaction.
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