# Solubility determination of Ca(OH)₂ in water

In one problem it's asking me to find the solubility of $\ce{Ca(OH)2}$ in water. In that problem it assumes the starting concentration $\ce{OH}$ to be zero. But for other problems like $\ce{BaSO4}$ in $0.1~M~~ \ce{Na2SO4}$, we account for the $\ce{SO4}$ in the $\ce{NaSO4}$. So in the first problem, why don't we worry about the $\ce{OH}$ from the self ionization of water? Is it just that extremely low of an amount?

• I don't get your point... It's like saying why we don't assume friction when computing the energy we spend to get a glass of water from the fridge. Never have we solved problems that talked about the true conditions in real life. – It's Over Mar 14 '15 at 19:58
• Yes, compared to the amount of $\ce{SO4}$ contributed by 0.1 M $\ce{Na2SO4}$, the amount of $\ce{OH^{-}}$ contributed by water is exceedingly small. – ron Mar 14 '15 at 20:16

The self ionization of $\ce{OH}$ is $10^{-7}$. If the concentration of $\ce{Ca(OH)2}$ is greater by a few magnitudes, then the self ionization should not matter much. In fact, $\ce{Ca(OH)2}$ is considered soluble, so if not in miniscule amounts, the self ionization should not matter much