To determine the pH of a 0.0040 M solution of calcium hydroxide, [tex]\( \text{Ca(OH)}_2 \)[/tex], follow the steps below:
1. Understand the dissociation of calcium hydroxide in water:
[tex]\[
\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2 \text{OH}^-
\][/tex]
Each molecule of [tex]\( \text{Ca(OH)}_2 \)[/tex] dissociates into one calcium ion ([tex]\( \text{Ca}^{2+} \)[/tex]) and two hydroxide ions ([tex]\( \text{OH}^- \)[/tex]).
2. Calculate the concentration of hydroxide ions ([tex]\( \text{OH}^- \)[/tex]):
Given the concentration of [tex]\( \text{Ca(OH)}_2 \)[/tex] is 0.0040 M, and knowing that each formula unit dissociates to produce two [tex]\( \text{OH}^- \)[/tex] ions, we can find the concentration of [tex]\( \text{OH}^- \)[/tex] as follows:
[tex]\[
\text{Concentration of } \text{OH}^- = 0.0040 \times 2 = 0.008 \text{ M}
\][/tex]
3. Calculate the pOH of the solution:
The pOH is related to the concentration of [tex]\( \text{OH}^- \)[/tex] ions and is calculated using the formula:
[tex]\[
\text{pOH} = -\log[\text{OH}^-]
\][/tex]
Substituting the concentration of [tex]\( \text{OH}^- \)[/tex]:
[tex]\[
\text{pOH} = -\log(0.008) \approx 2.097
\][/tex]
4. Determine the pH of the solution:
The pH and pOH are related by the equation:
[tex]\[
\text{pH} + \text{pOH} = 14
\][/tex]
Therefore, we can calculate the pH as follows:
[tex]\[
\text{pH} = 14 - \text{pOH}
\][/tex]
Substituting the value found for pOH:
[tex]\[
\text{pH} = 14 - 2.097 \approx 11.903
\][/tex]
Hence, the pH of the 0.0040 M solution of [tex]\( \text{Ca(OH)}_2 \)[/tex] is approximately 11.903.
