# Why does a 23*g mass of calcium hydroxide dissolved in a 400*mL volume of water, have [HO^-]-=1.55*mol*L^-1?

Oct 16, 2016

It seems that you have grasped the concept of stoichiometry:

$C a {\left(O H\right)}_{2} \left(s\right) \rightarrow C {a}^{2 +} + 2 H {O}^{-}$

#### Explanation:

$\left[H {O}^{-}\right]$ $=$ $\text{Moles of hydroxide ion"/"Volume of solution}$

$\text{Moles of calcium hydroxide} = \frac{23 \cdot g}{74.1 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.310 \cdot m o l$. $\text{Moles of calcium} = 0.155 \cdot m o l$.

$\left[H {O}^{-}\right]$ $=$ $\frac{2 \times 0.310 \cdot m o l}{0.400 \cdot L}$ $=$ $1.55 \cdot m o l \cdot {L}^{-} 1$, because each formula unit of $C a {\left(O H\right)}_{2}$ delivers one $C {a}^{2 +}$, but $2 \times H {O}^{-}$.

Each mole of calcium hydroxide clearly gives 2 moles of hydroxide anion upon dissolution, simply because of the formulation of calcium hydroxide. Is this clear?

To put it another way, the given solution is $1.55 \cdot m o l \cdot {L}^{-} 1$ with respect to $H {O}^{-}$, BUT half this concentration with respect to $C a {\left(O H\right)}_{2} \left(a q\right)$ and $C {a}^{2 +}$.