Problem 38 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

An aqueous solution is prepared by dissolving \(0.1\) mole \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in sufficient water to get \(100 \mathrm{ml}\) solution at \(25^{\circ} \mathrm{C}\). For \(\mathrm{H}_{2} \mathrm{CO}_{3}, \quad K_{\mathrm{a} 1}=4.0 \times 10^{-6}\) and \(K_{\mathrm{a} 2}=5.0 \times 10^{-11} .\) The only incorrect equilibrium concentration is (a) \(\left[\mathrm{H}^{+}\right]=6.32 \times 10^{-4} \mathrm{M}\) (b) \(\left[\mathrm{HCO}_{3}\right]=2 \times 10^{-3} \mathrm{M}\) (c) \(\left[\mathrm{CO}_{3}^{2-}\right]=5 \times 10^{-11} \mathrm{M}\) (d) \(\left[\mathrm{OH}^{-}\right]=5 \times 10^{-12} \mathrm{M}\)

Step-by-Step Solution

Answer

The incorrect equilibrium concentration is (a) [H+]=6.32 x 10^{-4} M.

Step 1: Calculate Initial Molarity of H2CO3

Convert the amount of moles of H2CO3 into molarity by dividing by the volume of the solution in liters. Since the volume is given in milliliters (ml), convert it to liters (L) by dividing by 1000. Molarity (M) = moles/volume(L).

Step 2: Write the Equilibrium Expressions

Write the equilibrium expressions for H2CO3 dissociation using the given acidity constants (Ka1 and Ka2).For the first dissociation: H2CO3 ⇌ H+ + HCO3- Ka1 = [H+][HCO3-]/[H2CO3]For the second dissociation: HCO3- ⇌ H+ + CO32-Ka2 = [H+][CO32-]/[HCO3-]

Step 3: Make Assumptions to Simplify Calculations

Assuming that the first dissociation is the dominant process and the concentration of H2CO3 remains approximately constant, the change in concentration for H+ and HCO3- will be almost equal and small compared to the initial concentration of H2CO3.

Step 4: Calculate [H+]

Using the Ka1 expression and assumptions: Ka1 = [H+]^2/[H2CO3] Solve for [H+]: [H+] = sqrt(Ka1 * [H2CO3])

Step 5: Calculate [HCO3-]

From the 1:1 stoichiometry of the first dissociation, [HCO3-] will be equal to [H+], which has already been calculated.

Step 6: Calculate [CO32-]

Using the Ka2 expression and the already calculated [H+] and assuming [HCO3-] remains largely unchanged, solve for [CO32-]: [CO32-] = Ka2 * [HCO3-] / [H+]

Step 7: Verify the Calculations

Check the calculated equilibrium concentrations against the given options to identify the incorrect value.

Step 8: Calculate [OH-]

Use the water ionization constant, Kw, to find [OH-]. At 25 degrees C, Kw is typically 1.0 x 10^-14. The [OH-] can be found with: [OH-] = Kw / [H+]

Step 9: Determine the Incorrect Concentration

By comparing the calculated concentrations with the given ones, determine which one is incorrect.

Key Concepts

Acid Dissociation Constants

When we talk about acid dissociation constants, we are referring to the strength of an acid in water. This strength is quantified by a value known as the acid dissociation constant, symbolized as Ka. It provides insight into the degree to which an acid donates its protons (H+) to the surrounding water molecules.

For a general acid dissociation reaction: \( HA \rightleftharpoons H^+ + A^- \) the Ka is calculated as:\[ K_a = \frac{[H^+][A^-]}{[HA]} \]In this equation, [H+] is the concentration of the hydrogen ion, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the undissociated acid. Higher values of Ka imply stronger acids, which dissociate more in water. Weak acids have smaller Ka values and do not dissociate completely.

Molarity Calculation

Molarity, often represented as M, is a unit of concentration for a solute in a solution. It is the number of moles of solute per liter of solution. The molarity calculation is a fundamental skill in chemistry, crucial for understanding solutions and reactions.

To calculate molarity, the formula used is:\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]When solving problems involving molarity, it's essential to remember to convert all volumes to liters if they are given in other units, such as milliliters, to maintain consistency.

Chemical Equilibrium

In chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, leading to a constant concentration of reactants and products in a closed system over time. It's a dynamic process—reactions are still occurring, but there is no net change in the amounts of substances.

The equilibrium constant, K, is used to describe the relationship between the concentrations of products and reactants. For a reaction:\[ aA + bB \rightleftharpoons cC + dD \]the equilibrium constant expression is:\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]This expression reflects the state of equilibrium and can help predict the direction of a reaction under different conditions. When working with weak acids or bases, an understanding of chemical equilibrium is crucial for accurate concentration calculations.

Aqueous Solution Chemistry

The study of aqueous solution chemistry is fundamental to understanding reactions in water. It involves the interaction of solutes (like acids, bases, and salts) with water—the solvent—to form a homogeneous mixture. The properties and behavior of ions and molecules in solution are governed by factors like concentration, temperature, and the presence of other substances.

Whether you are dealing with acid-base reactions, solubility equilibria, or redox reactions, the principles of aqueous solution chemistry will help you predict outcomes and calculate key parameters such as pH, pOH, and solubility product. It's a complex interplay, but mastering these concepts allows for a deeper understanding of chemical processes in an aqueous environment.