Can anyone help me calculate the mass concentration of $\ce{Ca(OH)2}$ solution with molar mass $\pu{74.1 g/mol}$ at the $\pu{pH of 12.8}$.

My work so far:

From the pH I get the molar concentration of $\pu{0.063 mol/L}$. Assuming we have $\pu{1L}$ of substance we have $\ce{n(OH-) = 2 \times n(CaOH2) = 0.126 mol/L}$

I put the results in $m = M \times n$ to get 9.35 g which is incorrect. Where did I go wrong?

  • $\begingroup$ you have $0.0315$ mol of $\ce{Ca(OH)2}$ when it dissociates completely $\endgroup$ – Adnan AL-Amleh Aug 16 '18 at 21:33
  • $\begingroup$ $m = 0.0315 \times 74.1$ to get 2.337 g $\endgroup$ – Adnan AL-Amleh Aug 16 '18 at 22:00
  • 1
    $\begingroup$ As a previous comment said, calcium concentration is half of hydroxide concentration. It yields $0.0315$ then your procedure is correct! $\endgroup$ – santimirandarp Aug 17 '18 at 4:11
  • $\begingroup$ Can someone please explain to me how the dissociation proceeds, so I understand how we arrive at the correct molar concentration $\endgroup$ – Bine Aug 18 '18 at 14:30
  • $\begingroup$ This is an ionic compound of$ Ca^{+2}$ ions and $OH^−$ ions. When an ionic compound dissolves, it separates into its constituent ions: $$Ca(OH)_2->Ca^{2+}_{(aq)}+2OH^−_{(aq)}$$ Because ${Ca(OH)_2}$ is a strong base, this reaction proceeds 100% to products. $\endgroup$ – Adnan AL-Amleh Aug 19 '18 at 19:05

Hint: (1) Properly, the pH tells you that you have $0.063$ molar hydroxide ion. (2) Calcium hydroxide, $\ce{Ca(OH)2}$, could dissociate more than once.

I'm not giving it all away, so good luck.

  • $\begingroup$ The OP already figured out that there were two hydroxides. That isn't the OP's problem... $\endgroup$ – MaxW Aug 16 '18 at 22:59
  • $\begingroup$ Not so, from the numerical calculations reported. The molar concentration was not given correctly. $\endgroup$ – Oscar Lanzi Aug 17 '18 at 0:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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