Question

IIClO is a weak acid. The concentration of II \(^{+}\) ions in \(0.1 \mathrm{M}\) solution of IIClO \(\left(K_{\mathrm{a}}=5 \times 10^{-5}\right)\) will be cqual to (1) \(7.07 \times 10^{-5} \mathrm{M}\) (2) \(5 \times 10^{-7} \mathrm{M}\) (3) \(5 \times 10^{-4} \mathrm{M}\) (4) \(7 \times 10^{-4} \mathrm{M}\)

Short answer

The concentration of II ions in the solution is \( 7.07 \times 10^{-4} \mathrm{M} \), which corresponds to option (4).

Step-by-step solution

Write the dissociation equation

Start by writing the dissociation equation for the weak acid IIClO:\[\text{IIClO} \rightleftharpoons \text{II}^+ + \text{ClO}^-\]

Use the Ka expression

The expression for the acid dissociation constant, Ka, is:\[K_a = \frac{[\text{II}^+][\text{ClO}^-]}{[\text{IIClO}]}\]Given, \[K_a = 5 \times 10^{-5}, [\text{IIClO}] = 0.1 \text{M}\]

Assume initial concentration

Let the concentration of II ions at equilibrium be x. Since the initial concentration of IIClO is 0.1 M, the concentration of IIClO at equilibrium will be reduced by x. Therefore,\[[\text{IIClO}] = 0.1 - x\]Given that IIClO is a weak acid, the value of x will be very small, so we assume that \[0.1 - x \approx 0.1\]

Apply the approximation in the Ka equation

The equation simplifies to:\[K_a = \frac{x^2}{0.1} = 5 \times 10^{-5}\]

Solve for x

Rearrange the equation to solve for x:\[x^2 = 5 \times 10^{-5} \times 0.1\]\[x^2 = 5 \times 10^{-6}\]\[x = \sqrt{5 \times 10^{-6}}\]\[x = 7.07 \times 10^{-4} \text{M}\]

Key concepts

weak acid

A weak acid is an acid that does not completely dissociate in water. Unlike strong acids, which fully dissociate, weak acids only partially break down into their ions. This means that in a solution, a significant amount of the weak acid remains in its undissociated form. Knowing whether an acid is weak or strong is important when calculating pH and for understanding equilibrium concentrations in a solution.

Ka expression

The acid dissociation constant, denoted as \(K_a\), is a measure of the strength of a weak acid. It represents the equilibrium constant for the dissociation of the acid into its ions in water. For an acid \(\text{HA}\), which dissociates as \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\), the \(K_a\) expression is given by:

• \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\)

A larger \(K_a\) value indicates a stronger weak acid (more dissociation), while a smaller \(K_a\) value indicates a weaker acid (less dissociation). In our exercise,\(K_a\) for IIClO is given as \(5 \times 10^{-5}\), showing it dissociates slightly in an aqueous solution.

equilibrium concentration

When a weak acid dissociates in water, it reaches an equilibrium where the rate of dissociation equals the rate of recombination. This equilibrium concentration can be calculated using the \(K_a\) value and the initial concentration of the acid. For example, in our problem, if the initial concentration of IIClO is 0.1 M, and \(x\) is the concentration of \(\text{II}^+\) ions at equilibrium, the equilibrium concentration of IIClO will be \(0.1 - x\). Because \(x\) is small, we approximate this as \(0.1\) to simplify calculations.

dissociation equation

Writing the dissociation equation is the first step in solving problems involving weak acids. For IIClO, the dissociation equation is:

• \(\text{IIClO} \rightleftharpoons \text{II}^+ + \text{ClO}^-\)

This equation shows that one molecule of IIClO dissociates into one ion of \(\text{II}^+\) and one ion of \(\text{ClO}^-\). With this, we set up the expression for \(K_a\) and solve for the equilibrium concentration of the ions. The detailed steps involve using the initial concentrations, applying approximations, and finally solving for \(x\) which represents the ion concentration at equilibrium. For IIClO, the concentration of the II ion turns out to be \(7.07 \times 10^{-4} \text{M}\).