Problem 85
Question
Consider a solution containing two weak monoprotic acids with dissociation constants \(K_{\mathrm{HA}}\) and \(K_{\mathrm{HB}}\). Find the charge balance equation for this system, and use it to derive an expression that gives the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) as a function of the concentrations of \(\mathrm{HA}\) and HB and the various constants.
Step-by-Step Solution
Verified Answer
The concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) in the solution, which is equivalent to the concentration of \([H^+]\), can be expressed as \([H^+] = \sqrt{K_{\mathrm{HA}}[HA] + K_{\mathrm{HB}}[HB]}\). This is derived by setting up the dissociation reactions, writing the charge balance equation, expressing the species in terms of equilibrium constants and substituting back into the charge balance equation.
1Step 1: Defining the Acid Dissociation Reactions
The acid dissociation of the weak monoprotic acids can be represented by the following reactions: \(HA \rightleftharpoons H^{+} + A^-\) and \(HB \rightleftharpoons H^{+} + B^- \). Both of these reactions have their respective equilibrium constants, \(K_{\mathrm{HA}}\) and \(K_{\mathrm{HB}}\), defined by concentration of products over concentration of reactants.
2Step 2: Formulate the Charge Balance Equation
In any given aqueous solution at equilibrium, the total positive charge must equal the total negative charge. Thus, the charge balance equation for this system can be written as: \([H^+] = [A^-] + [B^-]\) which indicates the amount of \(H^+\) ions equals the amount of \(A^-\) and \(B^-\) ions.
3Step 3: Define Concentrations Using Equilibrium Constants
We can use the equilibrium constants to express each species in terms of the others. The equilibrium expressions for \(HA\) and \(HB\) can be respectively written as \(K_{\mathrm{HA}} = \frac{[H^+][A^-]}{[HA]}\) and \(K_{\mathrm{HB}} = \frac{[H^+][B^-]}{[HB]}\). From these, we can derive expressions for the concentrations of \(A^-\) and \(B^-\). Thus, \([A^-] = \frac{K_{\mathrm{HA}}[HA]}{[H^+]}\) and \([B^-] = \frac{K_{\mathrm{HB}}[HB]}{[H^+]}\).
4Step 4: Substitute the Concentrations in the Charge Balance Equation
Substitute the derived expressions for \([A^-]\) and \([B^-]\) in the charge balance equation to obtain an expression for \([H^+]\) as a function of the concentrations of \(HA\) and \(HB\) and the various constants: \([H^+] = \frac{K_{\mathrm{HA}}[HA]}{[H^+]}+ \frac{K_{\mathrm{HB}}[HB]}{[H^+]}\). This equation can be rearranged to give \([H^+] = \sqrt{K_{\mathrm{HA}}[HA] + K_{\mathrm{HB}}[HB]}\).
Key Concepts
Weak AcidsEquilibrium ConstantsCharge Balance EquationHydronium Ion Concentration
Weak Acids
A weak acid is an acid that only partially dissociates in solution. In other words, it doesn't fully break apart into ions like a strong acid does. This partial dissociation means that in a weak acid solution, you will find a mix of ions and intact acid molecules. When an acid dissociates, it releases protons (\(H^+\)), which are responsible for the acidic properties of the solution.
Examples of weak acids include acetic acid (CH₃COOH), which you'll find in vinegar, and citric acid, commonly found in citrus fruits.
Examples of weak acids include acetic acid (CH₃COOH), which you'll find in vinegar, and citric acid, commonly found in citrus fruits.
- Partial dissociation affects the pH of the solution, making weak acids have higher pH values than strong acids for the same concentration.
- The equilibrium state of weak acids is characterized by a reversible reaction between the acid and its dissociated ions.
Equilibrium Constants
Equilibrium constants are vital quantities in chemistry that help us understand how reactions proceed in dynamic balance. For weak acids, the equilibrium constant, often denoted as\(K_a\), reflects the extent to which the acid dissociates into its ions. Specifically,\(K_a\) is defined as the ratio of the concentrations of the products to the reactants at equilibrium: \[K_a = \frac{[H^+][A^-]}{[HA]}\]
- A higher \(K_a\) value indicates a stronger acid because more acid molecules dissociate at equilibrium.
- A lower \(K_a\) value signifies a weaker acid with fewer dissociated ions.
Charge Balance Equation
The charge balance equation is a crucial concept when analyzing solutions at equilibrium. It states that in any solution, the total positive charge must equal the total negative charge, ensuring electrical neutrality.
For a solution with two weak acids, HA and HB, this balance is defined as \[[H^+] = [A^-] + [B^-]\]. This means the concentration of hydrogen ions \([H^+]\) matches the total concentration of the anions \([A^-]\) and \([B^-]\).
For a solution with two weak acids, HA and HB, this balance is defined as \[[H^+] = [A^-] + [B^-]\]. This means the concentration of hydrogen ions \([H^+]\) matches the total concentration of the anions \([A^-]\) and \([B^-]\).
- This equation allows us to connect the contributions of different species to the solution's charge balance.
- By ensuring all charges are balanced, we can predict and analyze the formation of ionic species in chemical reactions.
Hydronium Ion Concentration
The hydronium ion, \(H_3O^+\), is a central player in acid-base chemistry, acting as a measure of the acidity of a solution. In water, \(H^+\) ions naturally exist as hydronium ions because protons associate with water molecules.
To find the \([H_3O^+]\) concentration of a weak acid solution, chemists use the charge balance equation combined with equilibrium expressions. The derived formula for two weak acids, when simplified, ends up being: \[[H^+] = \sqrt{K_{\text{HA}}[HA] + K_{\text{HB}}[HB]}\]
To find the \([H_3O^+]\) concentration of a weak acid solution, chemists use the charge balance equation combined with equilibrium expressions. The derived formula for two weak acids, when simplified, ends up being: \[[H^+] = \sqrt{K_{\text{HA}}[HA] + K_{\text{HB}}[HB]}\]
- This expression indicates how the acid dissociation constants and initial concentrations influence the hydronium ion concentration.
- It provides insight into how mixed solutions of weak acids behave compared to single solutions.
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