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 $${\mathrm{HCO}}_3^{-}(aq)\rightleftharpoons {\mathrm{H}}^{+}(aq)+{\mathrm{CO}}_3^{2-}(aq)K=5.6\times {10}^{-11}$$ 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

The equilibrium concentration of ${\mathrm{CO}}_3^{2-}$ is approximately $3.0\times {10}^{-6}\text{M}$.

Step-by-step solution

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.

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: $$K=\frac{[{\mathrm{H}}^{+}][{\mathrm{CO}}_3^{2-}]}{[{\mathrm{HCO}}_3^{-}]}$$

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: $$[{\mathrm{HCO}}_3^{-}]=0.16-x\text{M}$$ $$[{\mathrm{H}}^{+}]=x\text{M}$$ $$[{\mathrm{CO}}_3^{2-}]=x\text{M}$$

4. Substitute the equilibrium concentrations into K expression

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

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 $3.0\times {10}^{-6}\text{M}$.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as "K," plays a crucial role in determining how far a reaction will proceed before reaching a state of balance. It is a numerical value that represents the ratio of the concentrations of products to reactants at equilibrium. For a given chemical reaction like the one in the exercise, the equilibrium equation can be used to express it. Given our equation:$${\mathrm{HCO}}_3^{-}(aq)\rightleftharpoons {\mathrm{H}}^{+}(aq)+{\mathrm{CO}}_3^{2-}(aq)$$The equilibrium constant expression for K is:$$K=\frac{[{\mathrm{H}}^{+}][{\mathrm{CO}}_3^{2-}]}{[{\mathrm{HCO}}_3^{-}]}$$Knowing "K" allows us to predict the concentrations of reactants and products at equilibrium. If "K" is very small, like in our example $5.6\times {10}^{-11}$, it indicates that the reaction heavily favors the reactants, meaning not much of them turn into products by the time equilibrium is reached. Which is why it's important in our calculations.

Concentration Calculations

Concentration calculations are an essential part of determining equilibrium in chemical reactions. These calculations involve converting the amount of substance, usually given in moles, into concentration, which is expressed in molarity (moles per liter, M) by dividing by the solution volume. For instance, we start with 0.16 moles of ${\mathrm{HCO}}_3^{-}$ in 1.00 L, which gives:$$[{\mathrm{HCO}}_3^{-}{]}_0=\frac{0.16\mathrm{mol}}{1.00\mathrm{L}}=0.16\mathrm{M}$$To find the concentrations at equilibrium, we use a method called the ICE table (Initial, Change, Equilibrium). Initially, the concentrations of ${\mathrm{H}}^{+}$ and ${\mathrm{CO}}_3^{2-}$ are zero because we only add ${\mathrm{HCO}}_3^{-}$ at the start. At equilibrium, assume "x" moles of ${\mathrm{HCO}}_3^{-}$ dissociate, leading to changes in the concentrations:

• $(0.16-x)\mathrm{M}$ for ${\mathrm{HCO}}_3^{-}$
• "x" M for ${\mathrm{H}}^{+}$
• "x" M for ${\mathrm{CO}}_3^{2-}$

These are inserted into the equilibrium expression to solve for "x," providing the equilibrium concentrations.

Reaction Quotient

The reaction quotient, "Q," serves a complementary role to the equilibrium constant in analyzing a chemical reaction's progress. While "K" provides insight into the concentrations at equilibrium, "Q" offers a snapshot of the system at any point before it reaches equilibrium. The expression for "Q" follows the same format as "K":$$Q=\frac{[{\mathrm{H}}^{+}][{\mathrm{CO}}_3^{2-}]}{[{\mathrm{HCO}}_3^{-}]}$$By comparing "Q" with "K," we can determine the direction in which a reaction will proceed:

• If $Q<K$, the reaction will move forward (to produce more products).
• If $Q>K$, the reaction will move in reverse (to produce more reactants).
• If $Q=K$, the system is at equilibrium.

In exercises, performing these calculations helps predict reaction behavior and verify if assumptions about reaction direction and equilibrium state hold true.