# Suppose 1.2 grams of calcium hydroxide are dissolved in 1.75 L of an aqueous solution. How do you determine the following: hydroxide ion concentration, hydronium ion concentration, calcium ion concentration, pH and poH?

Jun 1, 2017

$\text{Concentration"="Amount of stuff"/"Volume of solution}$

#### Explanation:

And generally, we report $\text{concentration}$ with units of $m o l \cdot {L}^{-} 1$.

And so $\left[C a {\left(O H\right)}_{2}\right] = \frac{\frac{1.2 \cdot g}{74.09 \cdot g \cdot m o {l}^{-} 1}}{1.75 \cdot L} = 9.26 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$.

Clearly, $\left[H {O}^{-}\right] = 2 \times 9.26 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$. Agreed? Because $\left[C a {\left(O H\right)}_{2}\right]$ speciates in solution to give $C {a}^{2 +}$, and 2 equiv of hydroxide ion.

And thus $\left[C {a}^{2 +}\right] = 9.26 \times {10}^{-} 3$, $\left[H {O}^{-}\right] = 1.85 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$.

$p O H$ is simply a measure of hydroxide ion concentration, and in aqueous solution we know that $p H + p O H = 14$.

$p O H = - {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} \left(1.85 \times {10}^{-} 2\right) = 1.73$

$p H = 14 - 1.73 = 12.27$.

For another example of this sort of problem, see here.