Problem 41 - Chemistry and Chemical Reactivity

Question

A 0.015 M solution of hydrogen cyanate, HOCN, has a pH of 2.67. (a) What is the hydronium ion concentration in the solution? (b) What is the ionization constant, \(K_{\mathrm{a}},\) for the acid?

Step-by-Step Solution

Answer

(a) The hydronium ion concentration is \(2.14 \times 10^{-3} \text{ M}\). (b) The ionization constant \(K_a\) is \(3.56 \times 10^{-4}\).

Step 1: Understanding the Relationship Between pH and Hydronium Ion Concentration

To find the hydronium ion concentration from the given pH value, we use the relationship: \( ext{pH} = -\log[\text{H}^+]\). In this case, the pH is 2.67. We'll use this to find \([\text{H}^+]\).

Step 2: Calculating Hydronium Ion Concentration

We rearrange the formula to find the concentration: \([\text{H}^+] = 10^{-\text{pH}}\). Substitute the pH value: \[[\text{H}^+] = 10^{-2.67} = 2.14 \times 10^{-3} \text{ M}.\] This is the concentration of hydronium ions in the solution.

Step 3: Expressing Ionization of HOCN

The acid ionizes in water as follows: \[\text{HOCN} \rightleftharpoons \text{H}^+ + \text{OCN}^-\]At equilibrium, the concentrations are: \([\text{H}^+] = [\text{OCN}^-] = x\) and \([\text{HOCN}] = 0.015 - x\).

Step 4: Using the Equilibrium Concentrations to Calculate Ka

Since \([\text{H}^+]\) is found to be \(2.14 \times 10^{-3} \text{ M}\), this is our \(x\). Substitute into the expression for the acid ionization constant \(K_a\):\[K_a = \frac{[\text{H}^+][\text{OCN}^-]}{[\text{HOCN}]} = \frac{x^2}{0.015 - x}.\]With \(x = 2.14 \times 10^{-3}\), \[K_a = \frac{(2.14 \times 10^{-3})^2}{0.015 - 2.14 \times 10^{-3}}\].

Step 5: Simplify and Calculate Ka

Calculate: \(x^2 = (2.14 \times 10^{-3})^2 = 4.58 \times 10^{-6}\) And the denominator: \(0.015 - 2.14 \times 10^{-3} = 0.01286\)Finally: \[K_a = \frac{4.58 \times 10^{-6}}{0.01286} = 3.56 \times 10^{-4}.\]This is the ionization constant \(K_a\) for the acid.

Key Concepts

Hydronium Ion Concentration

The hydronium ion concentration in a solution is a key measure that indicates how acidic a solution is. When an acid dissolves in water, it donates protons (H^+ ions) to water molecules, forming hydronium ions (H_3O^+).
This occurs because water acts as a base, accepting the hydrogen ion:H_2O + H^+ \rightarrow H_3O^+. This hydronium ion is what we actually measure when determining the acidity of a solution.

To find the hydronium ion concentration, we use the relationship with pH. The equation is given by:

• \[\text{pH} = -\log [\text{H}_3O^+]\]

This equation tells us that the pH is the negative logarithm of the hydronium ion concentration.
If we know the pH of a solution, we can reverse this calculation to find the hydronium ion concentration:

• \[[\text{H}_3O^+] = 10^{-\text{pH}}\]

In our exercise, with a pH of 2.67, the hydronium ion concentration is calculated as:

• \[[\text{H}_3O^+] = 10^{-2.67} = 2.14 \times 10^{-3} \text{ M}\]

This shows the amount of hydronium ions present in the solution, indicating its acidity.

pH and Ion Concentration

Understanding how pH relates to ion concentration is fundamental in chemistry. pH is a numerical representation of acidity or alkalinity in a solution.
The lower the pH, the more acidic the solution, and consequently, the higher the concentration of hydronium ions.

This is because the pH scale is logarithmic. Each whole number change on the scale corresponds to a tenfold change in ion concentration.
To simplify, here's how the pH scale relates to ion concentration:

• A pH of 7 is neutral, where [\text{H}^+] = [\text{OH}^-].
• pH less than 7 indicates an acidic solution, with more H^+ ions than OH^- ions.
• pH greater than 7 indicates a basic solution, with more OH^- ions.

In mathematical terms, the concentration of hydrogen ions at any given pH can be calculated using:

• \[[\text{H}^+] = 10^{-\text{pH}}\]

This simple equation illustrates the relationship between a solution's pH and its hydrogen ion concentration, helping us better understand the nature of any solution we're working with.

Equilibrium Expressions

When studying reactions in aqueous solutions, equilibrium expressions are crucial for understanding how the balance between reactants and products establishes over time.
In the context of acid ionization, equilibrium expressions allow us to calculate the ionization constant, K_a, of an acid.

The ionization constant is a measure of the strength of an acid, representing the acid's ability to donate protons in water.

• Strong acids have high K_a values, meaning they ionize extensively in solution.
• Weak acids have low K_a values, indicating less ionization.

For the ionization of an acid like hydrogen cyanate (HOCN), the equilibrium expression is:

• \[\text{HOCN} \rightleftharpoons \text{H}^+ + \text{OCN}^-\]
• At equilibrium, concentrations become [\text{H}^+] = [\text{OCN}^-] = x and [\text{HOCN}] = 0.015 - x.
• The K_a expression is K_a = \frac{[\text{H}^+][\text{OCN}^-]}{[\text{HOCN}]}.

This tells us the ratio of product concentrations to reactant concentrations at equilibrium.
By substituting known values, calculations can be made to find K_a, providing insight into the acidic behaviour of the compound in the solution.