Question

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

Answer 1

It seems that you have grasped the concept of stoichiometry:

`Ca(OH)_2(s) rarr Ca^(2+) + 2HO^-`

Explanation:

`[HO^-]` `=` `"Moles of hydroxide ion"/"Volume of solution"`

`"Moles of calcium hydroxide"=(23*g)/(74.1*g*mol^-1)` `=` `0.310*mol`. `"Moles of calcium"=0.155*mol`.

`[HO^-]` `=` `(2xx0.310*mol)/(0.400*L)` `=` `1.55*mol*L^-1`, because each formula unit of `Ca(OH)_2` delivers one `Ca^(2+)`, but `2xxHO^-`.

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*mol*L^-1` with respect to `HO^-`, BUT half this concentration with respect to `Ca(OH)_2(aq)` and `Ca^(2+)`.