Question

Calculate the \(\mathrm{pH}\) and the equilibrium concentration of \(\mathrm{HClO}\) in a \(0.10 \mathrm{M}\) solution of hypochlorous acid. \(K_{\mathrm{a}}\) \(\mathrm{HClO}=2.8 \times 10^{-8}\)

Short answer

The pH of the 0.10 M solution of hypochlorous acid is approximately 4.78, and the equilibrium concentration of HClO in the solution is approximately 0.0999833 M.

Step-by-step solution

Understand the dissociation reaction and set up the equilibrium expression

The dissociation reaction of hypochlorous acid (HClO) in water can be represented as: $$\mathrm{HClO} \rightleftharpoons \mathrm{H^{+}} + \mathrm{ClO^{-}}$$ Initially, we have 0.10 M HClO and no H+ or ClO- ions. At equilibrium, some of the HClO will dissociate into H+ and ClO- ions. We can represent the changes in concentration with the variable x: $$[\mathrm{H^{+}}] = x$$ $$[\mathrm{ClO^{-}}] = x$$ $$[\mathrm{HClO}] = 0.10 - x$$ Using the given Ka value, we can set up the equilibrium expression: $$K_{\mathrm{a}} = \frac{[\mathrm{H^{+}}][\mathrm{ClO^{-}}]}{[\mathrm{HClO}]}$$

Make an assumption to simplify the problem

We can assume that the change in concentration x is very small compared to the initial concentration of HClO (0.10 M). This means that \((0.10 - x) \approx 0.10\). By making this assumption, we can now write the equilibrium expression as: $$2.8 \times 10^{-8} = \frac{x \cdot x}{0.10}$$

Solve for x, which represents the equilibrium concentration of H+ ions

To solve for x, we need to solve the following equation: $$2.8 \times 10^{-8} = \frac{x^2}{0.10}$$ First, multiply both sides by 0.10 to isolate \(x^2\): $$x^2 = 2.8 \times 10^{-8} \times 0.10$$ Now, calculate the square root of both sides to find x: $$x = \sqrt{2.8 \times 10^{-8} \times 0.10}$$ $$x = 1.67 \times 10^{-5}$$ Here, x represents the equilibrium concentration of H+ ions and ClO- ions, which are both equal to \(1.67 \times 10^{-5} \mathrm{M}\).

Calculate the equilibrium concentration of HClO

To find the equilibrium concentration of HClO, we can now use the value of x from step 3: $$[\mathrm{HClO}]_{\mathrm{eq}} = 0.10 - x$$ $$[\mathrm{HClO}]_{\mathrm{eq}} = 0.10 - 1.67 \times 10^{-5}$$ $$[\mathrm{HClO}]_{\mathrm{eq}} = 0.0999833\mathrm{M}$$

Calculate the pH of the solution

The pH is defined as the negative logarithm of the H+ ion concentration: $$\mathrm{pH} = -\log {[\mathrm{H^{+}}]}$$ In this case, \([\mathrm{H^{+}}] = x = 1.67 \times 10^{-5} \mathrm{M}\). Thus, we can find the pH as follows: $$\mathrm{pH} = -\log (1.67 \times 10^{-5})$$ $$\mathrm{pH} \approx 4.78$$ The pH of the 0.10 M solution of hypochlorous acid is approximately 4.78, and the equilibrium concentration of HClO in the solution is approximately 0.0999833 M.