Question

Iodic acid, \(\mathrm{HIO}_{3}\), is an important oxidizing agent and a moderately strong acid. In a \(0.100 \mathrm{M}\) solution, \(\left[\mathrm{H}^{+}\right]=7.1\) \(\times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} .\) Calculate the \(K_{\mathrm{a}}\) and \(\mathrm{p} K_{\mathrm{a}}\) for iodic acid.

Short answer

The acid dissociation constant (Ka) for iodic acid is \(5.041 \times 10^{-3}\) and the pKa is 5.30.

Step-by-step solution

Identify the dissociation reaction of the acid

Write down the dissociation reaction for iodic acid, \(\mathrm{HIO}_3\), in water: \[\mathrm{HIO}_3(aq) \rightarrow \mathrm{H}^+(aq) + \mathrm{IO}_3^-(aq)\]

Write the expression for the acid dissociation constant (Ka)

The formula for the acid dissociation constant (Ka) for iodic acid is given as: \[K_a = \frac{[\mathrm{H}^+][\mathrm{IO}_3^-]}{[\mathrm{HIO}_3]}\] where the concentration of products and reactants are those at equilibrium.

Substitute given concentration values

Since the concentration of \(\mathrm{H}^+\) ions is given, assume the concentration of \(\mathrm{IO}_3^-\) ions is the same at equilibrium, and the concentration of undissociated \(\mathrm{HIO}_3\) can be approximated as the initial concentration since it is a weak acid: \[K_a = \frac{(7.1 \times 10^{-2})(7.1 \times 10^{-2})}{0.100}\]

Calculate the acid dissociation constant (Ka)

Perform the multiplication and division to find the value of \(K_a\): \[K_a = \frac{(7.1 \times 10^{-2})^2}{0.100} = 5.041 \times 10^{-3}\]

Calculate the pKa

The relationship between \(K_a\) and \(pK_a\) is given by: \[pK_a = -\log(K_a)\]. Substitute the calculated value of \(K_a\) into this formula to find \(pK_a\): \[pK_a = -\log(5.041 \times 10^{-3})\]

Evaluate the pKa

Using a calculator, determine the \(pK_a\) value: \[pK_a = -\log(5.041 \times 10^{-3}) = 2.30 - (-3) = 2.30 + 3 = 5.30\]

Key concepts

Acid Dissociation Constant

The acid dissociation constant (Ka) is a quantitative measure of an acid's strength—that is, its tendency to donate a proton (H+) in an aqueous solution. We can represent this property through a simple equilibrium expression. For a generic acid, HA, the dissociation in water can be written as:
\[\begin{equation} \text{HA(aq)} \rightleftharpoons \text{H}^+(aq) + \text{A}^-(aq) \tag{1} onumber \end{equation}\]The equilibrium constant for this reaction, denoted as Ka, is expressed as: \[\begin{equation} K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \tag{2} onumber \end{equation}\]where the concentrations are those at equilibrium. Concentrations of pure liquids and solids are not included in the equilibrium expression. A higher Ka value indicates a stronger acid, as it implies a greater concentration of H+ ions in solution. Understanding this concept is crucial for analyzing acid-base reactions in chemistry.

Ka and pKa Calculation

Calculating the Ka and pKa of an acid involves determining its equilibrium constant and then expressing its strength on a logarithmic scale, which offers more manageable numbers for comparison. The pKa provides a more straightforward way to ascertain an acid's strength. The formula to find pKa from Ka is as follows:
\[\begin{equation} pK_a = -\text{log}(K_a) \tag{3} onumber \end{equation}\]To calculate the pKa, thus, you first need the Ka value, which is computed by measuring the equilibrium concentrations of the ions and the undissociated acid. For iodic acid or any acid, this step can be made simpler if one recognizes that for weak acids, the initial concentration of the acid is roughly equal to the concentration of the acid at equilibrium. With the Ka determined, the pKa is just the negative logarithm of that value. A lower pKa value corresponds to a stronger acid, which may seem counterintuitive, but it's due to the inverse relationship with the logarithmic scale.

Acid-Base Equilibrium

Acid-base equilibrium forms the basis for understanding how acids and bases interact in solution and is governed by the principle of Le Chatelier. This principle states that a dynamic equilibrium will adjust to counteract any change imposed on it, thereby maintaining its constant state. In the context of acid dissociation, we refer to the balance between the undissociated acid and its ions in an aqueous solution. This equilibrium is not static but dynamic, meaning the reaction continuously proceeds in both the forward and reverse directions. Equilibrium is reached when the rate of the forward reaction, in which the acid donates H+ ions, equals the rate of the reverse reaction, where the conjugate base (A-) accepts H+ ions. A variety of factors can affect equilibrium, such as concentration changes, temperature, and the presence of other substances, and understanding this equilibrium is key to many practical applications in chemistry, from buffer solutions to pharmaceuticals.