Question

Solid calcium hydroxide, Ca(OH)2, is dissolved in water until the pH of the solution is 10.23. What is the concentration of calcium ion [Ca2+]?

Answer 1

Solid calcium hydroxide, $\mathrm{Ca(OH)}_2$, is dissolved in water until the pH of the solution is 10.23. What is the concentration of calcium ion $[\mathrm{Ca}^{2+}]$?

Solution:

In water $\mathrm{Ca(OH)}_2$ dissociates according to the equation:

$$\mathrm{Ca(OH)}_2 \leftrightarrow 2\mathrm{OH}^- + \mathrm{Ca}^{2+}$$

We assume that the pH of the water has been 7, after dissolved $\mathrm{Ca(OH)}_2$ in water the pH increased to 10.23. From this it follows that the concentration of ion $\mathrm{OH}^-$ increase by $10^{-3.23}\mathrm{mol/L}$ or $5.89\cdot 10^{-4}\mathrm{mol/L}$

$(\Delta pH = 10.23 - 7 = 3.23$, so $\Delta pH = \Delta pOH = 3.23$. Due to the fact that $pOH = -\lg [\mathrm{OH}^-]$, than $[\mathrm{OH}^-] = 10^{-3.23}\mathrm{M})$.

Because, during the dissociates of $\mathrm{Ca(OH)}_2$ the ions $[\mathrm{Ca}^{2+}]$ form twice less than the ions $[\mathrm{OH}^-]$, the concentration of $[\mathrm{Ca}^{2+}]$ is: $5.89\cdot 10^{-4}/2 = 2.945\cdot 10^{-4}\mathrm{mol/L}$.

**Answer**: $[\mathrm{Ca}^{2+}] = 2.945\cdot 10^{-4}\mathrm{mol/L}$