Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 77

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 \(0.16 \mathrm{~mol}\) 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 \(\mathrm{CO}_3^{2-}\) is approximately \(3.0 \times 10^{-6} \ \text{M}\).

Step by step solution

01

1. Write down the reaction and initial concentrations

The reaction can be written as: \[ \mathrm{HCO}_3^-(aq) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{CO}_3^{2-}(aq) \] Initially, we are given that 0.16 moles of \(\mathrm{HCO}_3^-\) is placed into 1.00 L of solution. We can convert moles into concentration by dividing the moles by the volume. Initial concentrations: \[ [\mathrm{HCO}_3^-]_0 = \frac{0.16 \ \text{mol}}{1.00 \ \text{L}} = 0.16 \ \text{M} \] As no other initial concentrations are provided, we can assume that the initial concentrations of \(\mathrm{H}^+\) and \(\mathrm{CO}_3^{2-}\) ions are zero.
02

2. Set up the equilibrium expression

The equilibrium constant for this reaction is given as \(K = 5.6 \times 10^{-11}\). We can write the equilibrium expression for K as follows: \[ K = \frac{[\mathrm{H}^+][\mathrm{CO}_3^{2-}]}{[\mathrm{HCO}_3^-]} \]
03

3. Define the changes in concentrations at equilibrium

At equilibrium, let 'x' moles of \(\mathrm{HCO}_3^-\) have dissociated. Thus, the change in concentration will be: \[ [\mathrm{HCO}_3^-] = 0.16 - x \ \text{M} \] \[ [\mathrm{H}^+] = x \ \text{M} \] \[ [\mathrm{CO}_3^{2-}] = x \ \text{M} \]
04

4. Substitute the equilibrium concentrations into K expression

Now, substitute the equilibrium concentrations into the expression for K: \[ 5.6 \times 10^{-11} = \frac{(x)(x)}{(0.16 - x)} \]
05

5. Solve the equation for x, which represents the equilibrium concentration of \(\mathrm{CO}_3^{2-}\)

To simplify solving the equation, we can make a simplifying assumption that \(x \ll 0.16\). This allows us to simplify the expression to: \[ 5.6 \times 10^{-11} \approx \frac{x^2}{0.16} \] Now, solve for x: \[ x^2 = 5.6 \times 10^{-11} \times 0.16 \] \[ x \approx \sqrt{5.6 \times 10^{-11} \times 0.16} \] \[ x \approx 3.0 \times 10^{-6} \] The assumption should be checked at this point: since \(3.0 \times 10^{-6} \ll 0.16\), the assumption holds true. So, the equilibrium concentration of \(\mathrm{CO}_3^{2-}\) is approximately \(3.0 \times 10^{-6} \ \text{M}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(1.604-\mathrm{g}\) sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas and \(6.400 \mathrm{~g}\) oxygen gas are sealed into a 2.50-L vessel at \(411^{\circ} \mathrm{C}\) and are allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium, the pressure of oxygen is \(0.326 \mathrm{~atm}\), and the pressure of water vapor is \(4.45\) atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.

Old-fashioned "smelling salts" consist of ammonium carbonate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) The reaction for the decomposition of ammonium carbonate $$ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ is endothermic. Would the smell of ammonia increase or decrease as the temperature is increased?

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\), consider two pos- sibilities: (a) you mix \(0.5\) mol of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or (b) you mix \(1.5 \mathrm{~mol}\) \(\mathrm{H}_{2}\) and \(0.5 \mathrm{~mol} \mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

The reaction $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ has \(K_{\mathrm{p}}=109\) at \(25^{\circ} \mathrm{C}\). If the equilibrium partial pressure of \(\mathrm{Br}_{2}\) is \(0.0159\) atm and the equilibrium partial pressure of \(\mathrm{NOBr}\) is \(0.0768\) atm, calculate the partial pressure of \(\mathrm{NO}\) at equilibrium.

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C}\), equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free