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 $$\mathrm{HCO}_{3}^{-}(a q) \leftrightharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) K=5.6 \times 10^{-11}$$ If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into 1.00 \(\mathrm{L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}^{2-2}\) ?

Short answer

The equilibrium concentration of \(CO_{3}^{2-}\) is approximately \(9.48 \times 10^{-7}\, M\).

Step-by-step solution

Write the given reaction and equilibrium constant

The reaction and equilibrium constant given are: \(HCO_{3}^{-}(aq) \leftrightharpoons H^{+}(aq) + CO_{3}^{2-}(aq)\;\;\;\;K = 5.6 \times 10^{-11}\)

Calculate initial concentrations

The initial concentration of bicarbonate ion is: \([HCO_{3}^{-}]_{initial} = \frac{0.16\, mol}{1.00\, L} = 0.16\, M\) Initially, there is no \(H^{+}\) or \(CO_{3}^{2-}\) from the reaction, so their concentrations are 0.

Create an ICE table

We can set up an ICE table (Initial, Change, Equilibrium) to find the equilibrium concentrations. The table looks like this: | | \(HCO_{3}^{-}\) | \(H^{+}\) | \(CO_{3}^{2-}\) | |-------|--------------|-------|--------------| |Initial| 0.16 | 0 | 0 | |Change | -x | +x | +x | |Equilibrium| 0.16-x | x | x | Where "x" represents the change in concentrations at equilibrium.

Write the equilibrium expression and substitute values

The equilibrium expression for the given reaction is: \(K = \frac{[H^{+}][CO_{3}^{2-}]}{[HCO_{3}^{-}]}\) Substitute the values from the ICE table: \(5.6 \times 10^{-11} = \frac{x\cdot x}{0.16-x}\)

Solve the equation for x

Assuming that x is much smaller than 0.16, we can simplify the equation as follows: \(5.6 \times 10^{-11} = \frac{x^2}{0.16}\) Now, solve for x: \(x^2 = 0.16\times (5.6\times 10^{-11}) \) \(x^2 = 8.96 \times 10^{-12}\) \(x = \sqrt{8.96 \times 10^{-12}}\) \(x = \approx 9.48 \times 10^{-7}\)

Determine the equilibrium concentration of \(CO_{3}^{2-}\)

The equilibrium concentration of \(CO_{3}^{2-}\) is equal to x. Therefore, \([CO_{3}^{2-}]_{equilibrium} \approx 9.48 \times 10^{-7}\, M\)

Key concepts

ICE Table

In chemical equilibrium problems, the ICE table is a critical tool used to analyze reactions. The acronym ICE stands for Initial, Change, and Equilibrium — which are the stages of concentration involved in these problems. To really understand how it works, imagine setting up a mini-database to track how the concentrations of reactants and products change over time.

Here's how it works:

• Initial: You start with the concentrations of the reactants and products before the reaction has proceeded (often, one or more may be zero).
• Change: This represents the amount the concentrations change as the system moves towards equilibrium; it's usually represented by a variable like "x".
• Equilibrium: By accounting for the change, you calculate the concentrations of all species at equilibrium.

Using the ICE table helps organize, visually track, and ultimately solve the calculations involved in chemical equilibria, such as determining how much of each species is present when the system is at equilibrium.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products in a reversible chemical reaction at equilibrium. This means that at equilibrium, the rate of the forward reaction equals the rate of the backward reaction, resulting in constant concentrations.

In the exercise, you have a reaction:\[ \mathrm{HCO}_{3}^{-} \leftrightharpoons \mathrm{H}^{+} + \mathrm{CO}_{3}^{2-} \] Initially, we only have bicarbonate ions (\(\mathrm{HCO}_{3}^{-}\)), and the equilibrium process starts with it breaking down into hydrogen (\(\mathrm{H}^{+}\)) and carbonate ions (\(\mathrm{CO}_{3}^{2-}\)).

At equilibrium, some amount of bicarbonate ions have decomposed, and thus the concentrations of all species must be calculated. It's these concentrations that the ICE table helps elucidate. The values at the equilibrium line represent the final equilibrium concentrations.

Equilibrium Constant

The equilibrium constant (\(K\)) is a crucial concept in determining the direction and extent of chemical reactions. It provides a quantifiable measure that expresses the ratio of concentrations of products to reactants at equilibrium.

In the problem, the equilibrium constant (\(K\)) is given as \(5.6 \times 10^{-11}\). A very small \(K\) value, like this one, indicates that the equilibrium heavily favors reactants over products. In a broader sense, it means that under equilibrium conditions, there will be much more starting material (reactant) than products.

Using the ICE table, you can substitute these values into the equilibrium expression:\[ K = \frac{[\mathrm{H}^{+}][\mathrm{CO}_{3}^{2-}]}{[\mathrm{HCO}_{3}^{-}]} \] and solve for the unknown concentration x (representing \([\mathrm{CO}_{3}^{2-}]\) at equilibrium). The size of \(K\) dictates how these concentrations adjust, emphasizing just how few carbonate ions are made relative to the other species in the chemical reaction.