Question

What is the pH of the aqueous solution of Ca(OH)2 with a calcium ion concentration of 6.33×10−3 M at 25°C?

Answer 1

To find the pH of the aqueous solution of \( \text{Ca(OH)}_2 \), we first determine the hydroxide ion concentration produced by the dissociation of \( \text{Ca(OH)}_2 \).

The dissociation equation for calcium hydroxide is:

\[
\text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2 \text{OH}^-
\]

From this equation, we see that one mole of \( \text{Ca(OH)}_2 \) produces one mole of \( \text{Ca}^{2+} \) ions and two moles of \( \text{OH}^- \) ions.

Given the calcium ion concentration \( [\text{Ca}^{2+}] = 6.33 \times 10^{-3} \, \text{M} \), this means that the concentration of hydroxide ions will be:

\[
[\text{OH}^-] = 2 \times [\text{Ca}^{2+}] = 2 \times 6.33 \times 10^{-3} \, \text{M} = 1.266 \times 10^{-2} \, \text{M}
\]

Next, we can calculate the pOH of the solution using the hydroxide ion concentration:

\[
\text{pOH} = -\log([\text{OH}^-]) = -\log(1.266 \times 10^{-2})
\]

Using a calculator:

\[
\text{pOH} \approx 1.895
\]

Now we can find the pH using the relation between pH and pOH:

\[
\text{pH} + \text{pOH} = 14
\]

So,

\[
\text{pH} = 14 - \text{pOH} \approx 14 - 1.895 \approx 12.105
\]

Therefore, the pH of the aqueous solution of \( \text{Ca(OH)}_2 \) is approximately:

\[
\boxed{12.1}
\]