Problem 3 - Chemical Principles in the Laboratory

Question

The pH of a 0.10 M HOBr solution is 4.8. a. What is \(\left[\mathrm{H}^{+}\right]\) in that solution? b. What is \(\left[\mathrm{OBr}^{-}\right] ?\) What is [HOBr]? (Where do the \(\mathrm{H}^{+}\) and \(\mathrm{OBr}^{-}\) ions come from?) c. What is the value of \(K_{\mathrm{a}}\) for HOBr? What is the value of \(\mathrm{pK}_{\mathrm{a}} ?\)

Step-by-Step Solution

Answer

[H⁺]=1.58×10^{-5} M, [OBr⁻]=1.58×10^{-5} M, [HOBr]≈0.10 M, Kₐ≈2.50×10^{-9}, pKₐ≈8.60.

Step 1: Calculate [H⁺]

Given the pH of the solution is 4.8, we use the formula to find the concentration of hydrogen ions \([\mathrm{H}^+]\): \([\mathrm{H}^+]=10^{-\text{pH}}\). Substituting the values, we get \([\mathrm{H}^+]=10^{-4.8}\). Calculating this gives \([\mathrm{H}^+]=1.58\times10^{-5}\) M.

Step 2: Identify Equilibrium Concentrations

In the dissociation of HOBr, \(\text{HOBr} \rightleftharpoons \text{H}^+ + \text{OBr}^-\). Initially, \([\text{HOBr}] = 0.10\) M. At equilibrium, \([\text{HOBr}]=0.10 - x\), \([\text{H}^+]=x\), \([\text{OBr}^-]=x\). Since \([\text{H}^+]=1.58\times10^{-5}\) M (from Step 1), \(x = 1.58\times10^{-5}\), thus \([\text{OBr}^-]=1.58\times10^{-5}\) M and \([\text{HOBr}]=0.10-1.58\times10^{-5}\approx0.10\) M.

Step 3: Calculate Kₐ for HOBr

The expression for the acid dissociation constant \(K_a\) is: \(K_a = \frac{[\text{H}^+][\text{OBr}^-]}{[\text{HOBr}]}\). Substituting the equilibrium concentrations: \(K_a = \frac{(1.58\times10^{-5})(1.58\times10^{-5})}{0.10}\approx2.50\times10^{-9}\).

Step 4: Calculate pKₐ for HOBr

To find \(\text{pK}_a\), use the relation \(\text{pK}_a = -\log(K_a)\). Substituting the value of \(K_a\): \(\text{pK}_a = -\log(2.50\times10^{-9})\approx8.60\).

Key Concepts

pH Calculation

Understanding pH Calculation is fundamental in acid-base chemistry. It represents the acidity or alkalinity of a solution. pH is calculated using the formula:

• \[pH = -\log_{10} [\text{H}^+]\]

It measures the concentration of hydrogen ions, \([ ext{H}^+]\), in a solution.
In our example with 0.10 M HOBr, the given pH is 4.8.
Using the formula \([ ext{H}^+] = 10^{-\text{pH}}\), we substitute the known pH value:
\([ ext{H}^+] = 10^{-4.8}\).
Calculating this gives:

• \[[ ext{H}^+] = 1.58\times10^{-5}\] M.

This concentration helps to understand how acidic the solution is, which is slightly above neutral.

Equilibrium Concentration

Determining Equilibrium Concentrations is crucial for understanding the ratios of reactants and products in a chemical equilibrium. For the dissociation of HOBr in our example:

• \[\text{HOBr} \rightleftharpoons \text{H}^+ + \text{OBr}^-\]

Initially, the concentration of HOBr is 0.10 M.
At equilibrium, we use the principle that the change in concentration for \(\text{H}^+\) and \(\text{OBr}^-\) is equal.
With our previously found hydrogen ion concentration \(1.58 \times 10^{-5}\) M,

• \([ ext{OBr}^-] = [\text{H}^+] = 1.58 \times 10^{-5}\) M

Since \(x\) is very small compared to 0.10 M, \([ ext{HOBr}]\) remains approximately \(0.10~M\).
The almost unchanged concentration of HOBr at equilibrium is a sign of a weak acid.

Acid Dissociation Constant (Ka)

The Acid Dissociation Constant, \(K_a\), is a measure of the strength of an acid in solution.
It quantifies the equilibrium concentrations between the acid and its ions.
For HOBr, the equilibrium expression is:

• \[K_a = \frac{[\text{H}^+][\text{OBr}^-]}{[\text{HOBr}]}\]

Using our calculated concentrations:

• \[[\text{H}^+] = 1.58\times10^{-5}\] M
• \[[\text{OBr}^-] = 1.58\times10^{-5}\] M
• \[[\text{HOBr}] \approx 0.10 M\]

Plug these values into the expression:

• \[K_a \approx \frac{(1.58\times10^{-5})(1.58\times10^{-5})}{0.10}\approx2.50\times10^{-9}\]

A low \(K_a\) indicates a weak acid, reflecting that HOBr does not completely ionize in the solution.

pKa Calculation

Calculating pKa provides insight into the acid's strength.
It is derived from the acid dissociation constant, \(K_a\).
The formula relating \(K_a\) and pKa is:

• \[pK_a = -\log_{10}(K_a)\]

Substituting our previously calculated \(K_a\) of \(2.50\times10^{-9}\):

• \[pK_a = -\log_{10}(2.50\times10^{-9})\approx8.60\]

A high pKa value indicates that HOBr is a weak acid because it resists ionizing in solution.
Understanding the relationship between \(K_a\) and pKa helps chemists and students predict how an acid behaves in different environments.