Question

Calculate the \(\left[\mathrm{OH}^{-}\right]\) of each aqueous solution with the following \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]:\) a. coffee, \(1.0 \times 10^{-5} \mathrm{M}\) b. soap, \(1.0 \times 10^{-8} \mathrm{M}\) c. cleanser, \(5.0 \times 10^{-10} \mathrm{M}\) d. lemon juice, \(2.5 \times 10^{-2} \mathrm{M}\)

Short answer

[OH^-] for coffee: 1.0 x 10^-9 M; soap: 1.0 x 10^-6 M; cleanser: 2.0 x 10^-5 M; lemon juice: 4.0 x 10^-13 M

Step-by-step solution

- Identify the relationship

The relationship between the concentration of hydronium ions \(\text{[H}_3\text{O}^+\)\text{)} and hydroxide ions \(\text{[OH}^-\)\text{)} in water is given by the ion-product constant for water, \(K_w \), which is \([H_3O^+][OH^-] = 1.0 \times 10^{-14} M^2\).

- Rearrange the equation

Solve for the hydroxide ion concentration, \(\text{[OH}^-\)\text{)} by rearranging the equation: \[ [OH^-] = \frac{K_w}{[H_3O^+]} \]

- Calculate [OH^-] for coffee

Given \(\text{[H}_3\text{O}^+\)\text{)} = \(1.0 \times 10^{-5} \text{M}\): \[ [OH^-] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-5}} = 1.0 \times 10^{-9} \text{M} \]

- Calculate [OH^-] for soap

Given \(\text{[H}_3\text{O}^+\)\text{)} = \(1.0 \times 10^{-8} \text{M}\): \[ [OH^-] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-8}} = 1.0 \times 10^{-6} \text{M} \]

- Calculate [OH^-] for cleanser

Given \(\text{[H}_3\text{O}^+\)\text{)} = \(5.0 \times 10^{-10} \text{M}\): \[ [OH^-] = \frac{1.0 \times 10^{-14}}{5.0 \times 10^{-10}} = 2.0 \times 10^{-5} \text{M} \]

- Calculate [OH^-] for lemon juice

Given \(\text{[H}_3\text{O}^+\)\text{)} = \(2.5 \times 10^{-2} \text{M}\): \[ [OH^-] = \frac{1.0 \times 10^{-14}}{2.5 \times 10^{-2}} = 4.0 \times 10^{-13} \text{M} \]

Key concepts

ion-product constant for water

Water dissociates into hydronium ions \(\text{[H}_3\text{O}^+\text{]}\) and hydroxide ions \[ \text{[OH}^- \text{]}\]. This occurs in both pure water and aqueous solutions. The ion-product constant for water, \[ K_w \], represents this equilibrium state and is a very small number: \[ K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14} \text{M}^2 \]. \ This constant value is crucial for understanding how the concentrations of \[ \text{[H}_3\text{O}^+\text{]} \] and \[ \text{[OH}^- \text{]} \] are related in water and solutions. Understanding \[ K_w \] allows us to calculate the concentration of either ion if we know the value of the other.

hydronium ion concentration

Hydronium ions are represented as \[ \text{[H}_3\text{O}^+\text{]} \] and are formed when water acts as an acid and donates a proton to a water molecule. This concentration influences the acidity of a solution. \ For example, a higher \[ \text{[H}_3\text{O}^+\text{]} \], such as \[ 2.5 \times 10^{-2} \text{M} \] in lemon juice, indicates a more acidic solution. In each of the given problems – coffee, soap, cleanser, and lemon juice – knowing the hydronium ion concentration lets us find out how much hydroxide is also present. \ This is essential for calculating pH and understanding the acidic or basic nature of the solution.

hydroxide ion concentration

Hydroxide ions, \[ \text{[OH}^- \text{]} \], form when water acts as a base and accepts a proton. If we know \[ \text{[H}_3\text{O}^+\text{]} \], we can calculate \[ \text{[OH}^- \text{]} \] using the formula \[ [OH^-] = \frac{K_w}{[H_3O^+]} \]. This relationship is key because it lets us understand the balance between acidic and basic properties in the solution. \ In the exercises, to find \[ [OH^-] \] for coffee \[ (\text{[H}_3\text{O}^+\text{]} = 1.0 \times 10^{-5} \text{M}) \], we use \[ K_w = 1.0 \times 10^{-14} \text{M}^2 \] and find \[ [OH^-] = 1.0 \times 10^{-9} \text{M} \]. This step-by-step calculation illustrates how closely the concentrations of these ions are linked.

pH and pOH relationship

The pH of a solution measures its acidity or basicity. It is defined as \[ \text{pH} = -\text{log}[H_3O^+] \]. Similarly, pOH measures the concentration of hydroxide ions: \[ \text{pOH} = -\text{log}[OH^-] \]. Importantly, for any aqueous solution at 25°C, the sum of pH and pOH equals 14: \[ \text{pH} + \text{pOH} = 14 \]. \ This relationship helps us easily switch between understanding the acidity and basicity of a solution. For example, if you calculate that a solution has a pH of 5, you immediately know the pOH is 9. This direct connection between pH and pOH through their logarithmic scales simplifies many calculations and helps in quick assessments of solution properties.