Question

Consider the reaction: $$\begin{array}{c}{\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)} \\\ {K_{\mathrm{c}}=1.8 \times 10^{-5} \mathrm{at} 25^{\circ} \mathrm{C}}\end{array}$$ If a solution initially contains \(0.210 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2},\) what is the equi- librium concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) at \(25^{\circ} \mathrm{C}\) ?

Short answer

The equilibrium concentration of \(H_3O^+\) at 25°C is \(1.94 \times 10^{-3} M\).

Step-by-step solution

Write the expression for the equilibrium constant

Use the provided chemical reaction to write the expression for the equilibrium constant, Kc. For the reaction \(HC_2H_3O_2(aq) + H_2O(l) \Longrightarrow H_3O^+(aq) + C_2H_3O_2^-(aq)\), the expression for Kc is \[K_{c} = \frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}\] where the concentrations of the products are in the numerator and the concentration of the reactant is in the denominator.

Set up the ICE table

Create an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations throughout the reaction. Initial concentrations are given. For the reaction there's no change in water concentration since it's a liquid. Assume that the change in the concentration for the products is +x and for the reactant is -x.| Reactant/Products | Initial (M) | Change (M) | Equilibrium (M) ||--------------------|-------------|------------|-------------------------|| HC2H3O2 | 0.210 | -x | 0.210 - x || H2O | - | - | - || H3O+ | 0 | +x | x || C2H3O2- | 0 | +x | x |

Write the Kc expression with equilibrium concentrations

Substitute the equilibrium concentrations from the ICE table into the Kc expression. The new Kc expression is \[1.8 \times 10^{-5} = \frac{x( x)}{(0.210 - x)}\] Solve this equation by assuming that \(x\) is much less than 0.210, so that the denominator will be approximately equal to the initial concentration, simplifying the equation to \[1.8 \times 10^{-5} = \frac{x^2}{0.210}\]

Solve for x

Multiply both sides by 0.210 and take the square root to solve for x. \[x^2 = 1.8 \times 10^{-5} \times 0.210\] \[x^2 = 3.78 \times 10^{-6}\] \[x = \sqrt{3.78 \times 10^{-6}}\] \[x = 1.94 \times 10^{-3}\]

Verify the assumption

Check that the value of x is indeed much less than 0.210 to validate the assumption made in Step 3. Here, \(1.94 \times 10^{-3}\) is much smaller than 0.210, so the assumption is valid.

Report the equilibrium concentration of H3O+

The equilibrium concentration of \(H_3O^+\) is equal to x. Therefore, the equilibrium concentration of \(H_3O^+\) at 25°C is \(1.94 \times 10^{-3} M\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rates of the forward reaction (reactants converting to products) and the reverse reaction (products converting to reactants) are equal, leading to no net change in the concentration of reactants and products over time. This balance between opposing reactions is dynamic, which means that even though the concentrations remain constant, the reactions continue to occur at a molecular level.

Understanding equilibrium is crucial in chemical analysis because many reactions in nature and industry reach a point where this balance is maintained. To characterize the point of equilibrium, we use the equilibrium constant, denoted as Kc for reactions in solution, which provides a ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients.

ICE Table Method

The ICE table method is a systematic approach to solving equilibrium problems in chemistry. ICE stands for Initial, Change, Equilibrium, which are the three states we evaluate in a chemical reaction. Here's how each part is used:

• Initial: These are the starting concentrations or pressures of reactants and products in the reaction.
• Change: This represents the change in concentration or pressure as the system moves towards equilibrium. It is often represented by '+x' for products and '-x' for reactants, assuming the reaction proceeds to form more products.
• Equilibrium: These values reflect the concentrations or pressures when the system is at equilibrium. They are calculated by combining the initial amounts and the change.

An ICE table helps organize the known and unknown variables in a reaction to solve for equilibrium concentrations. It is particularly useful when some components, like solids and pure liquids, do not appear in the equilibrium expression since their concentrations do not change.

Acid Dissociation Constant

The acid dissociation constant, Ka, is a specific type of equilibrium constant that measures the strength of an acid in solution. It is defined as the ratio of the concentration of the dissociated (ionized) form of the acid to the undissociated form. For a generic weak acid, HA, dissociating in water to form H+ and A-, the expression for Ka is:
\[ K_{a} = \frac{[H^+][A^-]}{[HA]} \]
Strong acids have large Ka values, indicating that they dissociate completely in solution, whereas weak acids have small Ka values and only partially dissociate. Calculating the acid dissociation constant involves determining the equilibrium concentrations of the acid and its respective ions. This concept is essential in understanding the behavior of acids and bases in solution and plays a crucial role in pH calculations, buffer solutions, and many aspects of analytical chemistry.