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?

  • $\begingroup$ 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. $\endgroup$ – M.A.R. Mar 14 '15 at 19:58
  • 1
    $\begingroup$ 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. $\endgroup$ – 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.