Question

Calculate the \(\mathrm{pH}\) and the concentrations of all species present \(\left(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{OH}, \mathrm{HIO}_{3}, \mathrm{IO}_{3}^{-}\right)\) in \(0.0500 \mathrm{M} \mathrm{HIO}_{3} . K_{a}\) for \(\mathrm{HIO}_{3}\) is \(1.7 \times 10^{-1}\)

Short answer

pH ≈ 1.39; [H3O+] ≈ 0.0405 M; [IO3-] ≈ 0.0405 M; [HIO3] ≈ 0.0095 M; [OH-] ≈ 2.47 × 10^-13 M.

Step-by-step solution

Write the Dissociation Reaction

The dissociation reaction of \text{HIO}_3 in water is given by: \[\text{HIO}_3 + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{IO}_3^-\] Here, \(\text{HIO}_3\) dissociates to produce \(\text{H}_3\text{O}^+\) and \(\text{IO}_3^-\).

Set up the Expression for the Acid Dissociation Constant

The acid dissociation constant \(K_a\) for \(\text{HIO}_3\) is given by: \[K_a = \frac{[\text{H}_3\text{O}^+][\text{IO}_3^-]}{[\text{HIO}_3]}\] Given that \(K_a = 1.7 \times 10^{-1}\) and the initial concentration of \(\text{HIO}_3\) is 0.0500 M.

Define the Change in Concentrations

Assume \(x\) is the change in concentration at equilibrium. Initially, the concentration of \(\text{H}_3\text{O}^+\) and \(\text{IO}_3^-\) is 0. At equilibrium, \[[\text{H}_3\text{O}^+] = x, \quad [\text{IO}_3^-] = x, \quad [\text{HIO}_3] = 0.0500 - x\]

Solve for x Using the Quadratic Equation

Substitute the equilibrium concentrations into the \(K_a\) expression:\[1.7 \times 10^{-1} = \frac{x^2}{0.0500 - x}\]Solve this equation to find \(x\). Since this \(K_a\) is relatively large, we should not assume \(0.0500 - x \approx 0.0500\). This requires solving the quadratic equation:\[x^2 + 1.7x - 0.085 = 0\]Use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute \(a = 1\), \(b = 1.7\), \(c = -0.085\):\[x = \frac{-1.7 \pm \sqrt{(1.7)^2 - 4(1)(-0.085)}}{2(1)}\]This gives \(x \approx 0.0405\).

Calculate \(\mathrm{pH}\) and Concentrations

Now that \(x\) is found to be 0.0405,- \([\text{H}_3\text{O}^+] = x \approx 0.0405 \text{ M}\)- \([\text{IO}_3^-] = x \approx 0.0405 \text{ M}\)- \([\text{HIO}_3] = 0.0500 - x = 0.0095 \text{ M}\)Calculate \(\mathrm{pH}\):\[\text{pH} = -\log[\text{H}_3\text{O}^+] = -\log(0.0405) \approx 1.39\]

Calculate Concentration of OH^-

Use the water autoionization constant \(K_w = 1.0 \times 10^{-14}\):\[K_w = [\text{H}_3\text{O}^+][\text{OH}^-]\]\[[\text{OH}^-] = \frac{1.0 \times 10^{-14}}{[\text{H}_3\text{O}^+]}\]Substitute \([\text{H}_3\text{O}^+]=0.0405\):\[[\text{OH}^-] = \frac{1.0 \times 10^{-14}}{0.0405} = 2.47 \times 10^{-13} \text{ M}\]

Key concepts

pH Calculation

When dealing with acids, understanding the concept of pH is essential. The pH scale is a measure of acidity or basicity in a solution. It's a logarithmic scale that runs from 0 to 14.
In our example, pH is determined from the concentration of hydronium ions (\[\text{H}_3\text{O}^+\]). The formula to calculate pH is:

• \[\text{pH} = -\log{[\text{H}_3\text{O}^+]}\]

This means that a higher concentration of \[\text{H}_3\text{O}^+\] results in a lower pH, indicating a stronger acidic solution.
For instance, in the solution with \[\text{HIO}_3\] that dissociates to form \[\text{H}_3\text{O}^+\], we calculated \[\text{H}_3\text{O}^+\] to be 0.0405 M. Plugging this into the formula gives us a pH of approximately 1.39. This describes a strongly acidic environment, as pH values below 7 indicate acidity. Short recap:

• Acids have pH values less than 7.
• The more hydronium ions, the lower the pH value.

Acid-Base Equilibrium

Acid-base equilibrium describes the balance between the concentrations of acids and bases in a solution. When acids dissolve in water, they dissociate, releasing ions that establish equilibrium in the solution.
For the reaction \[\text{HIO}_3 + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{IO}_3^-\], the molecule \[\text{HIO}_3\] releases hydronium ions (\[\text{H}_3\text{O}^+\]) and iodate ions (\[\text{IO}_3^-\]).
This balance is described by the acid dissociation constant (\[K_a\]):

• \[K_a = \frac{[\text{H}_3\text{O}^+][\text{IO}_3^-]}{[\text{HIO}_3]}\]

The value of \[K_a\] provides insight into the strength of an acid. A higher \[K_a\] indicates a stronger acid as it shows a greater extent of dissociation. In our exercise, \[K_a\] is given as\[1.7 \times 10^{-1}\], signifying significant dissociation, which supports the low pH found.
Key points:

• Equilibrium implies the rate of acid dissociation equals the rate of association back into the acid form.
• Changes in concentration affect the equilibrium, driven by Le Chatelier's Principle.

Quadratic Formula in Chemistry

The quadratic formula is a handy tool in chemistry, especially when dealing with equilibrium problems that involve squared terms. This comes into play when the acid dissociates, and the equilibrium expression involves a term squared.
In our solution for \[\text{HIO}_3\] dissociation, we derived the equation:

• \[1.7 \times 10^{-1} = \frac{x^2}{0.0500 - x}\]

This leads to the quadratic form:

• \[x^2 + 1.7x - 0.085 = 0\]

Using the quadratic formula to solve: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] helps us find possible values for \[x\], the concentration at equilibrium. Using values \[a = 1\], \[b = 1.7\], and \[c = -0.085\], we found \[x\approx 0.0405\].
This is used to determine concentrations of equilibrium species. It is crucial when the simplification of assumptions (like neglecting \[x\] in the denominator) can't be applied.
Highlight:

• The quadratic formula provides solutions in scenarios where the concentration changes are significant and assumptions aren't valid.
• Understanding its use helps solve complex equilibrium problems in chemistry.