Question

You have \(75.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HA. After adding \(30.0 \mathrm{mL}\) of \(0.10 M \mathrm{NaOH},\) the \(\mathrm{pH}\) is \(5.50 .\) What is the \(K_{\mathrm{a}}\) value of \(\mathrm{HA} ?\)

Short answer

The Ka value of the weak acid HA is approximately \(2.08 \times 10^{-5}\).

Step-by-step solution

Calculate initial moles of HA and NaOH

Calculate the initial moles of HA and NaOH before the reaction. For HA: moles = Molarity × Volume moles of HA = 0.10 M × 75.0 mL = 0.10 M × 0.075 L = 0.0075 moles For NaOH: moles = Molarity × Volume moles of NaOH = 0.10 M × 30.0 mL = 0.10 M × 0.030 L = 0.0030 moles

Perform the reaction between HA and NaOH and determine the remaining moles of HA

Since NaOH is a strong base, it will react completely with the weak acid HA. Determine the remaining moles of HA after the reaction. moles of HA remaining = initial moles of HA - moles of NaOH = 0.0075 - 0.0030 = 0.0045 moles

Calculate the final volume of the solution

Determine the final volume of the solution by adding the initial volumes of both HA and NaOH solutions. Final volume = initial volume of HA + initial volume of NaOH = 75.0 mL + 30.0 mL = 105.0 mL = 0.105 L

Calculate the equilibrium concentration of HA

Determine the equilibrium concentration of HA by dividing its remaining moles by the final volume of the solution. \[C_{HA, eq} = \frac{moles\:of\:HA\: remaining}{final\:volume} = \frac{0.0045\: moles}{0.105\: L} = 0.0429 M\]

Use the pH to find the equilibrium concentration of H+ ions

We are given that the pH at equilibrium is 5.50. Use the formula: pH = -log10[H+] Solve for [H+]: \[[H^{+}] = 10^{-pH} = 10^{-5.50} = 3.16 \times 10^{-6} M\]

Determine the equilibrium concentration of A- ions

Since the moles of NaOH completely reacted with HA, the moles of A- ions produced at equilibrium are the same as the initial moles of NaOH. Calculate the equilibrium concentration of A- ions by dividing the moles of A- ions by the final volume of the solution. \[C_{A^-, eq} = \frac{moles\:of\:A^-}{final\:volume} = \frac{0.0030\: moles}{0.105\: L} = 0.0286 M\]

Calculate the Ka value using equilibrium concentrations

Use the ion concentrations at equilibrium to find the Ka value of HA using the formula: \[K_{a} = \frac{[H^{+}][A^{-}]}{[HA]_{eq}}\] Substitute the equilibrium concentrations found in previous steps: \[K_{a} = \frac{(3.16 \times 10^{-6})(0.0286)}{0.0429} = 2.08 \times 10^{-5}\] The Ka value of the weak acid HA is approximately \(2.08 \times 10^{-5}\).

Key concepts

Equilibrium Concentration

In any chemical reaction that reaches equilibrium, the concentration of reactants and products becomes stable over time.
The equilibrium concentration is an essential aspect of acid-base equilibria.
It helps understand how much of an acid or base remains unreacted or has been converted during the reaction.

To find the equilibrium concentration of HA in our problem, we first need to know how much of the initial acid remains after reacting with the base.
We calculate it by determining the initial moles of the acid (HA) and the base (NaOH).
Once the base has completely reacted with the acid, the remaining amount of HA can be measured by subtracting moles that reacted from the initial.

• In this solution, 0.0075 moles of HA were reduced by 0.0030 moles of NaOH, leaving 0.0045 moles of HA.
• We then adjust for the final volume of the solution — the original volumes of both reactants combined make 0.105 L.

The equilibrium concentration of HA, therefore, is calculated by dividing these remaining moles by the total solution volume.
Understanding these steps provides clarity on how a solution's concentration is influenced by a neutralization reaction.

pH Calculation

The pH of a solution is a measure of its acidity or alkalinity, indicating the concentration of hydrogen ions ([H⁺]) present.
The pH is calculated as the negative logarithm of the hydrogen ion concentration, expressed in the formula: - pH = -log10[H⁺].
For instance, in the given exercise, we have the pH of 5.50.
This means that the [H⁺] can be found using [H⁺] = 10^{-5.50}.

• This calculation involves using exponential functions, which turn a negative exponent into a positive but small fraction.
• The resulting hydrogen ion concentration, [H⁺] = 3.16 × 10^{-6} M, confirms the acidity of the solution.

Each change in pH reflects a 10-fold change in [H⁺], with lower pH values indicating more acidic solutions.
It's crucial for understanding the properties of the solution, especially in acid-base reactions where pH directly represents equilibrium conditions and the success of neutralizing reactions.

Acid Dissociation Constant (Ka)

The Acid Dissociation Constant, (Ka), is a fundamental concept in understanding the strength of an acid in solution.
It quantifies how much an acid dissociates into its ions at equilibrium.
In the context of the exercise, Ka can be measured using the equilibrium concentrations of ions produced in the reaction:

• The formula is Ka = \([H⁺][A⁻] / [HA]_{eq}\), showing the ratio of the products of the dissociation to the unreacted acid.

By substituting the concentration values previously calculated, \([H⁺] = 3.16 × 10^{-6} M\), \([A⁻] = 0.0286 M\), and \([HA]_{eq} = 0.0429 M\), we achieve the Ka = 2.08 × 10^{-5} for HA.

This small value indicates a weak acid as it dissociates minimally in a solution.
Outlining this calculation provides practical insight into the nature and behavior of acids in various conditions, which can be crucial for predicting reactions and behaviors in more complex chemical systems.