Question

Calculate the pH of a \(0.010-M\) solution of iodic acid (HIO \(_{3}, K_{\mathrm{a}}\) \(=0.17 )\)

Short answer

The pH of a 0.010-M solution of iodic acid (HIO3) with \(K_a = 0.17\) is approximately 1.38.

Step-by-step solution

Write the chemical equation

Begin by writing the chemical equation for the ionization of iodic acid in water: \[HIO_3 (aq) + H_2O (l) \rightleftharpoons H_3O^+(aq) + IO_3^-(aq)\]

Write the equilibrium expression

Next, write the equilibrium expression (Ka) for the ionization of iodic acid in water: \[K_a = \frac{[H_3O^+][IO_3^-]}{[HIO_3]}\]

Define the equilibrium concentrations

Let x be the equilibrium concentration of H3O⁺ and IO3⁻ ions. The initial concentration of iodic acid is 0.010 M, so the equilibrium concentrations can be defined as: - [H3O⁺] = x - [IO3⁻] = x - [HIO3] = 0.010 - x

Substitute equilibrium concentrations into the Ka expression

Substitute the equilibrium concentrations in the Ka expression: \[0.17 = \frac{x^2}{0.010 - x}\]

Solve the quadratic equation for x

Since Ka is significantly larger than x, we can approximate the equation by neglecting x in the denominator: \[0.17 = \frac{x^2}{0.010}\] Now, solve for x: \[x^2 = 0.17 \times 0.010\] \[x^2 = 0.0017\] \[x = \sqrt{0.0017}\] \[x = 0.0412\]

Calculate the pH of the solution

Now that we have the concentration of H3O⁺ ions, we can calculate the pH of the solution using the formula: pH = -log[H3O⁺] \[pH = -log(0.0412)\] \[pH \approx 1.38\] The pH of a 0.010-M solution of iodic acid (HIO3) is approximately 1.38.

Key concepts

Equilibrium Expression

Understanding the equilibrium expression is crucial to solving acid dissociation problems, such as calculating the pH of iodic acid. When an acid like HIO_{3} dissolves in water, it establishes an equilibrium between its dissociated ions and the undissociated acid molecules.

The equilibrium expression provides a mathematical representation of this equilibrium state. For iodic acid, the chemical equation for its dissociation is:- \[ HIO_3(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + IO_3^-(aq) \]The equilibrium constant expression, denoted as \( K_a \), is written in terms of the concentrations of these species:- \[ K_a = \frac{[H_3O^+][IO_3^-]}{[HIO_3]} \]This equation tells us the ratio of the products of the reaction to the reactants at equilibrium.

Equilibrium expressions enable us to quantify how far a reaction proceeds and help predict the concentrations of ions in solution once equilibrium is established.

Ionization Constant

The ionization constant, commonly referred to as the \( K_a \) value, is a measure of the strength of an acid in water. It tells us how completely an acid ionizes or dissociates in solution.

In the case of iodic acid, the given \( K_a \) is 0.17, indicating it is a moderately strong acid. - Stronger acids have higher \( K_a \) values, reflecting a greater propensity to donate protons (H⁺) to water molecules, forming hydronium ions (H₃O⁺).- Weaker acids have lower \( K_a \) values, suggesting less dissociation in water.

When solving problems involving acid dissociation, the \( K_a \) value is essential in setting up the equilibrium expression and helps determine the concentrations of ions at equilibrium. It allows us to predict the amount of the acid that has dissociated and the resulting pH of the solution. This is why having a correct and clear understanding of the ionization constant is so important for pH-related calculations.

Quadratic Approximation

When dealing with equilibrium problems involving acids, solving the equilibrium expression often leads to a quadratic equation. However, sometimes we can simplify the mathematics through a method known as quadratic approximation.

In this exercise, the approximation was applied in step 5. Given that the ionization constant \( K_a \) for iodic acid (0.17) is relatively high, we assume that the change \( x \) in concentration of the acid isn't too large compared to its initial concentration. - This allows us to approximate the expression \( 0.010 - x \approx 0.010 \) in the denominator of the equation.- The equation simplifies significantly, making it easier to solve for \( x \), which represents the concentration of hydronium ions, \([H_3O^+]\).

Quadratic approximation is a helpful mathematical tool in chemistry, enabling us to solve complex equations with greater ease, provided certain conditions are met. This method makes calculations accessible without compromising the accuracy significantly, particularly when \( K_a \) is either very small or relatively large.

Acid Dissociation

Acid dissociation involves the process where an acid donates a proton to water, forming hydronium ions and conjugate base ions. For iodic acid, this is represented by the reaction:- \[ HIO_3(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + IO_3^-(aq) \]The degree of dissociation determines the acidity of the solution, directly impacting its pH.

Understanding acid dissociation helps grasp why a solution of iodic acid is acidic and how we can calculate the resulting pH. The extent of dissociation is predicted using the ionization constant \( K_a \), which reflects the strength of an acid. - A high \( K_a \) value, such as 0.17 for iodic acid, implies a significant degree of dissociation, producing more hydronium ions and lowering the pH. - Conversely, a very low \( K_a \) value indicates incomplete dissociation, resulting in a higher pH.
The concept of acid dissociation is fundamental in chemistry, explaining how acids behave in solutions and the underlying reason for pH variations in different substances.