Question

What are \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in a \(6.02 \times 10^{-4} \mathrm{M}\) solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) ?

Short answer

The \([\text{H}^+]\) is \(8.31 \times 10^{-12}\, \text{M}\) and \([\text{OH}^-]\) is \(1.204 \times 10^{-3}\, \text{M}\).

Step-by-step solution

Determine the Concentration Increase from Dissociation

Recognize that calcium hydroxide, \(\text{Ca(OH)}_2\), dissociates into one \(\text{Ca}^{2+}\) ion and two \(\text{OH}^-\) ions in solution. Therefore, the concentration of \(\text{OH}^-\) ions will be twice that of the original \(\text{Ca(OH)}_2\) concentration. Calculate the \(\text{OH}^-\) concentration as follows: \[ [\text{OH}^-] = 2 \times 6.02 \times 10^{-4} \text{ M} = 1.204 \times 10^{-3} \text{ M} \]

Apply the Water Ion Product Constant

Use the relation between hydrogen and hydroxide ion concentrations given by the water ion product constant: \[ K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \text{ at 25°C} \]Substitute the \(\text{OH}^-\) concentration and solve for \([\text{H}^+]\): \[ [\text{H}^+] = \frac{1 \times 10^{-14}}{1.204 \times 10^{-3}} = 8.31 \times 10^{-12} \text{ M} \]

Verify Solution with pH and pOH

Calculate \(\text{pOH}\) using the relation \(\text{pOH} = -\log[\text{OH}^-]\): \[ \text{pOH} = -\log(1.204 \times 10^{-3}) = 2.92 \]Then calculate \(\text{pH}\) using the relation \(\text{pH} = 14 - \text{pOH}\): \[ \text{pH} = 14 - 2.92 = 11.08 \]Check that the calculated pH corresponds to the calculated \([\text{H}^+]\).

Key concepts

Hydroxide Ion Concentration

When calcium hydroxide, \({\text{Ca(OH)}_2}\), dissolves in water, it splits into one calcium ion \(\text{Ca}^{2+}\) and two hydroxide ions \(\text{OH}^-\). This means that for every mole of \({\text{Ca(OH)}_2}\) dissolved, two moles of \(\text{OH}^-\) ions are produced. This is an essential concept, as it helps us determine the concentration of hydroxide ions in the solution.
In this example, you start with a \(6.02 \times 10^{-4} \, \text{M}\) solution of calcium hydroxide. Since \(\text{Ca(OH)}_2\) produces two hydroxide ions for every formula unit that dissociates, the concentration of \(\text{OH}^-\) is twice that. Therefore, the hydroxide ion concentration is calculated as follows:

• The hydroxide ion concentration is \[ [\text{OH}^-] = 2 \times 6.02 \times 10^{-4} \, \text{M} = 1.204 \times 10^{-3} \, \text{M} \]

Understanding this relationship is crucial for determining the solution's alkalinity and further calculations involving pH and pOH.

Water Ion Product Constant

The water ion product constant, symbolized as \(K_w\), is a crucial concept in chemistry that represents the equilibrium constant for the ionization of water.
At 25°C, water ionizes very slightly to create equal concentrations of hydrogen ions \(\text{H}^+\) and hydroxide ions \(\text{OH}^-\). The equilibrium expression for this ionization is expressed as:

• \[ K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \]

This constant allows us to relate the concentrations of hydrogen and hydroxide ions together in any aqueous solution. By knowing the concentration of one of these ions, you can calculate the concentration of the other.
In the given problem, using the hydroxide ion concentration \(1.204 \times 10^{-3} \, \text{M}\) allows us to find the hydrogen ion concentration with the equation:

• \[ [\text{H}^+] = \frac{1 \times 10^{-14}}{1.204 \times 10^{-3}} = 8.31 \times 10^{-12} \, \text{M} \]

The constancy of \(K_w\) in normal conditions is a fundamental aspect of acid-base chemistry.

pH and pOH Calculation

Calculating pH and pOH is a vital part of understanding the acidity or basicity of a solution. These values help scientists and students alike to determine how acidic or basic a solution is using a clear and straightforward scale.
#### Understanding pOH Calculations
pOH is calculated using the formula:

• \[ \text{pOH} = -\log[\text{OH}^-] \]
• For \(\text{OH}^- = 1.204 \times 10^{-3} \, \text{M}\), the calculation is:
  \[ \text{pOH} = -\log(1.204 \times 10^{-3}) = 2.92 \]

#### Converting pOH to pH
Remember the handy relationship between pH and pOH:

• \[ \text{pH} + \text{pOH} = 14 \]

Knowing this, we can find the pH:

• \[ \text{pH} = 14 - \text{pOH} = 14 - 2.92 = 11.08 \]

The measured pH confirms that the solution is basic, aligning neatly with the high concentration of hydroxide ions. This straightforward approach helps ensure all calculations are on point and the solution's chemical nature is well-understood.