Question

Hypochlorous acid, HClO, has a p \(K_{a}\) of \(7.54 .\) What are \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right], \mathrm{pH},\left[\mathrm{ClO}^{-}\right],\) and \([\mathrm{HClO}]\) in \(0.115 \mathrm{M} \mathrm{HClO} ?\)

Short answer

\[ [\text{H}_3\text{O}^+] = 5.75 \times 10^{-5} \text{ M} \], \[ \text{pH} = 4.24 \], \[ [\text{ClO}^-] = 5.75 \times 10^{-5} \text{ M} \], \[ [\text{HClO}] = 0.115 \text{ M} \]

Step-by-step solution

- Write the ionization equation

HClO ionizes in water as follows: \[ \text{HClO} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{ClO}^- \]

- Set up the expression for the acid dissociation constant (Ka)

The expression for the acid dissociation constant is given by: \[ K_a = \frac{[\text{H}_3\text{O}^+][\text{ClO}^-]}{[\text{HClO}]} \]Given that the pKa is 7.54, first convert it to Ka using the relation:\[ K_a = 10^{-pKa} = 10^{-7.54} \]

- Calculate the value of Ka

Calculate Ka: \[ K_a = 10^{-7.54} = 2.88 \times 10^{-8} \]

- Set up the initial concentrations

Initially, the concentration of HClO is 0.115 M, and the concentrations of H3O+ and ClO- are both approximately 0.

- Define the change in concentrations at equilibrium

Assume that x M of HClO dissociates:\[ [\text{HClO}] = 0.115 - x \]\[ [\text{H}_3\text{O}^+] = x \]\[ [\text{ClO}^-] = x \]Substituting these into the expression for Ka gives:\[ K_a = \frac{x \times x}{0.115 - x} \]

- Solve the quadratic equation

Since Ka is very small, we can approximate that:\[ 0.115 - x \thickapprox 0.115 \]Thus, the expression simplifies to:\[ 2.88 \times 10^{-8} = \frac{x^2}{0.115} \]Solving for x:\[ x^2 = (2.88 \times 10^{-8}) \times 0.115 \]\[ x^2 = 3.31 \times 10^{-9} \]\[ x = \frac{\text{H}_3\text{O}^+}{\text{ClO}^-} = \text{ [H}_3\text{O}^+] \thickapprox 5.75 \times 10^{-5} \]

- Calculate the pH

The pH is given by the negative logarithm of the \text{H}_3\text{O}^+ concentration:\[ \text{pH} = -\text{log}([\text{H}_3\text{O}^+]) \]\[ \text{pH} = -\text{log}(5.75 \times 10^{-5}) = 4.24 \]

- Calculate the equilibrium concentrations

[HClO] = Initial concentration - x\[ [\text{HClO}] = 0.115 - 5.75 \times 10^{-5} \]\[ [\text{HClO}] \thickapprox 0.115 \][ClO-] = x\[ [\text{ClO}^-] = 5.75 \times 10^{-5} \]

Key concepts

pKa calculation

Let's start by understanding what pKa is. pKa is a measure of the strength of an acid. It tells us how easily an acid can donate a proton (H+ ion) in water. The formula to calculate Ka from pKa is simple: \[ K_a = 10^{-pKa} \] In the given exercise, the pKa of hypochlorous acid (HClO) is 7.54. Plugging that into our formula, we get: \[ K_a = 10^{-7.54} \times 10^{-8} = 2.88 \times 10^{-8} \] This value represents how likely the acid (HClO) is to dissociate in water.

acid dissociation constant (Ka)

The acid dissociation constant, Ka, is an important figure in understanding acid-base chemistry. It tells us the equilibrium constant for the dissociation reaction of the acid in water. For HClO, the dissociation equation is:
\[ \text{HClO} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{ClO}^- \] In this reaction, HClO donates a proton to water, forming H3O+ (hydronium ion) and ClO^- (hypochlorite ion). The Ka expression for this reaction is: \[ K_a = \frac{[\text{H}_3\text{O}^+][\text{ClO}^-]}{[\text{HClO}]} \] It provides a quantitative measure of the position of the equilibrium. A higher Ka value means a stronger acid, since it dissociates more in water.

pH calculation

Now, let's calculate the pH. The pH measures the hydrogen ion concentration, [H3O+], in a solution and is given by the formula: \[ \text{pH} = -\text{log}([\text{H}_3\text{O}^+]) \] From the steps, we established that: \[ [\text{H}_3\text{O}^+] = 5.75 \times 10^{-5} \] To find the pH, we just take the negative logarithm of this concentration: \[ \text{pH} = -\text{log}(5.75 \times 10^{-5}) = 4.24 \] Thus, the pH of the 0.115 M HClO solution is about 4.24.

equilibrium concentrations

Finally, we need to find the equilibrium concentrations of the different species in the solution. At equilibrium, we assume that x M of HClO has dissociated. We use the following relations:

• \text{Initial concentration of HClO} = 0.115 M
• \text{Change in concentration} = x

Therefore, the equilibrium concentrations are:

• \text{[HClO]} = 0.115 - x ≈ 0.115 M (since x is very small)
• \text{[ClO]^-} = x = 5.75 \times 10^{-5} M

Using Ka, we find: \[ K_a = \frac{x \times x}{0.115} = 2.88 \times 10^{-8} \] Solving for x, we get the concentration of H3O+, which in this case equals: \[ x = 5.75 \times 10^{-5} \] This shows that at equilibrium, we have approximately 0.115 M HClO and 5.75 \times 10^{-5} M of both H3O+ and ClO^-.