Question

A solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) has a measured \(\mathrm{pH}\) of 11.60 . What is the molar concentration of the \(\mathrm{Ca}(\mathrm{OH})_{2}\) in the solution? What is the molar concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) if the solution is diluted so that the \(\mathrm{pH}\) is \(10.60 ?\)

Short answer

The original molar concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(\frac{10^{-2.40}}{2}\) M, and after dilution to a pH of 10.60, it is \(\frac{10^{-3.40}}{2}\) M.

Step-by-step solution

Convert pH to pOH

First, calculate the pOH from the given pH by using the relationship: pOH = 14 - pH. For the original pH of 11.60, the pOH will be 14 - 11.60 = 2.40.

Calculate the hydroxide ion concentration

Next, use the pOH to find the hydroxide ion (OH-) concentration. This is done by the equation: [OH-] = 10^{-pOH}. With a pOH of 2.40, [OH-] = 10^{-2.40}.

Calculate the molarity of Ca(OH)2

Since \(\mathrm{Ca}(\mathrm{OH})_{2}\) dissociates into one calcium ion and two hydroxide ions, the molarity of \(\mathrm{Ca}(\‌‌‌​mathrm{OH})_{2}\) is half the molarity of OH-. Using the equation, [Ca(OH)2] = [OH-]/2, we can find the concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Calculate pOH after dilution

When the pH is changed to 10.60 due to dilution, calculate the new pOH which is 14 - 10.60 = 3.40.

Calculate the new hydroxide ion concentration

Using the new pOH, calculate the hydroxide ion concentration: [OH-] = 10^{-3.40}.

Calculate the molarity of Ca(OH)2 after dilution

Finally, find the new molarity of \(\mathrm{Ca}(\mathrm{OH})_{2}\) by halving the calculated hydroxide ion concentration post-dilution: [Ca(OH)2] = [OH-]/2.

Key concepts

pH and pOH Relationship

Understanding the relationship between pH and pOH is crucial when tackling acid-base chemistry problems. pH is a measure of the hydrogen ion concentration in a solution, while pOH gives us information about the hydroxide ion concentration. They are interconnected by a simple mathematical relationship: the sum of the pH and pOH of a solution at 25°C (298 K) always equals 14.

This is expressed by the equation: \begin{align*} \text{pH} + \text{pOH} = 14 \end{align*}This equation is derived from the ion product constant for water, which at room temperature is \(1.0 \times 10^{-14}\).When you know either the pH or pOH of a solution, finding the other is a simple calculation. For instance, if a solution has a pH of 3, its pOH would be \(14 - 3 = 11\). If the pH increases, indicating a decrease in hydrogen ion concentration, the pOH decreases correspondingly, showing an increase in hydroxide ion concentration.

Hydroxide Ion Concentration Calculation

The calculation of hydroxide ion concentration when given the pOH is straightforward thanks to the relationship \begin{align*} [\text{OH}^-] = 10^{-\text{pOH}} \end{align*}Let's say a solution has a pOH of 2.4. We can calculate the hydroxide ion concentration by raising 10 to the power of the negative pOH: \([\text{OH}^-] = 10^{-2.4}\).Why does this work? The 'p' in pOH (and pH) stands for 'negative logarithm of', and so pOH is the negative logarithm of the hydroxide ion concentration. Inverting this relationship by using the antilogarithm (or the ten raised to the power of the negative pOH), gives us the actual concentration of the hydroxide ions. For acids, bases, and neutral solutions, understanding and calculating hydroxide ion concentration is essential for analyzing properties like acidity, alkalinity, and pH levels.

Dilution and pH Change

The process of dilution involves adding more solvent to a solution, which lowers the concentration of the solute. In the context of acids and bases, when a basic solution like \(\text{Ca}(\text{OH})_2\) is diluted, the concentration of \(\text{OH}^-\) ions decreases. This affects the pH and pOH of the solution. A handy rule to remember is that when diluting a solution, the number of moles of solute remains the same but the volume increases, thereby reducing the molarity or concentration.

After dilution, the pH of the solution will decrease, making it less basic (or more acidic). Conversely, the pOH will increase. Given the inverse relationship between pH and pOH, as the pOH goes up due to the lowered concentration of hydroxide ions, the pH goes down. This relationship holds true until the solution becomes so dilute that it approximates the neutrality of pure water, a pH of 7.