Problem 89 - Chemistry An Atoms First Approach

Question

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 \(\mathrm{CO}_{3}{\underline{\phantom{xx}}}^{2-}\) ?

Step-by-Step Solution

Answer

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

Step 1: 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

Step 2: 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 |

Step 3: 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: \(5.6 \times 10^{-11} = \frac{x \cdot x}{0.16 - x}\)

Step 4: 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: \(x \approx \sqrt{(5.6 \times 10^{-11})(0.16)}\) \(x \approx 3.003 \times 10^{-6}\)

Step 5: 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.

Key Concepts

ICE Table

The ICE table is a vital tool in understanding chemical equilibrium. By using it, you can maintain a clear overview of how concentrations of reactants and products shift over time. It is structured into rows for Initial, Change, and Equilibrium concentrations of reactants and products.

- **Initial (I):** This row displays the initial concentrations before the reaction reaches equilibrium.
- **Change (C):** Here, you track the alteration in concentrations as the reaction moves toward equilibrium. Usually, these changes are noted with "+x" or "-x" to indicate gain or loss.
- **Equilibrium (E):** This row indicates the concentrations at equilibrium, combining initial concentrations and changes.

For example, in the reaction \[\mathrm{HCO}_{3}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq)+\mathrm{CO}_{3}^{2-}(aq),\]using an ICE table helps track the decrease in \(\mathrm{HCO}_{3}^{-}\) and the increase in \(\mathrm{H}^{+}\) and \(\mathrm{CO}_{3}^{2-}\). This clear format helps you solve equilibrium problems systematically and logically.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a critical component in understanding the balance between reactants and products at equilibrium. It quantifies how far a reaction will proceed toward products at equilibrium.

- **Expression:** For a general reaction \[aA + bB \rightleftharpoons cC + dD,\]the equilibrium constant expression is \[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}.\]
- **Interpretation:** A large \(K\) value means the reaction favors products at equilibrium, whereas a small \(K\) signifies that reactants are favored.

In the context of our reaction \[\mathrm{HCO}_{3}^{-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{CO}_{3}^{2-},\]with a \(K = 5.6 \times 10^{-11}\), it indicates that the reaction greatly favors the reactants, hence a tiny amount of the products \(\mathrm{H}^{+}\) and \(\mathrm{CO}_{3}^{2-}\) form. Using the \(K\) value alongside the ICE table, you can predict how much of each species will be present at equilibrium.

Concentration Calculations

Calculating concentrations in equilibrium reactions involves a systematic approach using the data available. In chemical equilibrium problems, after setting up an ICE table, you incorporate these values into the equilibrium constant expression to solve for unknown concentrations.

### Solving for Equilibrium Concentrations1. **Write the equilibrium expression:** Start with the balanced chemical equation to set up your expression correctly.
2. **Substitute ICE values:** Plug the equilibrium concentrations from the ICE table into the expression.
3. **Simplify Approximations:** If \(K\) is very small, it suggests little product formation, allowing approximations like \((a - x) \approx a\). This simplifies calculations greatly.
4. **Solve the Equation:** For small \(K\) like \(5.6 \times 10^{-11}\), constant algebraic steps can resolve x, indicating minor product concentration.

For the reaction, substituting and solving gives the concentration of \(\mathrm{CO}_{3}^{2-}\) as \(3.003 \times 10^{-6}\) M at equilibrium. These calculations empower learners to connect theoretical aspects with quantitative insights in chemical reactions.