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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?

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  • $\begingroup$ 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
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    $\begingroup$ As a previous comment said, calcium concentration is half of hydroxide concentration. It yields $0.0315$ then your procedure is correct! $\endgroup$ – user43021 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
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There is a general law about the solubility of salts containing doubly or triply charged ions, which is not often presented in textbooks. The law is the following : The measured solubility of such a salt or such a hydroxide is ALWAYS greater than the calculated values derived from solubility products, or from pH measurements. This is due to the fact that the solubility product and the pH data do not take into account the existence of intermediate ions.

In the case of $\ce{Ca(OH)2}$, there are of course ions like $\ce{Ca^2+}$ and $\ce{OH-}$ in its solution. But there are also ions like $\ce{Ca(OH)+}$, and non dissociated molecules $\ce{Ca(OH)2}$. Alas these ions and molecules are rarely mentioned in usual textbooks. But this explains why the solubility of $\ce{Ca(OH)2}$ is bigger than the values obtained from pH measurements. I know that the average reader is reluctant to believe me. I may be more persuasive if I use another example.

Let's speak about the solubility of $\ce{H3PO4}$. Any student in chemistry knows that the solubility is not related to the solubility product $\ce{[H+]^3[PO4^3-]}$, because of the existence of intermediated ions, like $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$. These ions are accepted in the large chemistry family. Everybody accepts that the solubility of $\ce{H3PO4}$ is much bigger that the value derived from the solubility products, or from pH measurements.

That is the same for $\ce{Ca(OH)2}$ and all substances containing a doubly charged ion like $\ce{CaSO4}$, etc. And here the activity is not sufficient to explain this discrepancy. I know that some chemists think that the activity may be introduced to explain this discrepancy. But in the case of $\ce{CaSO4}$ for example, the concentration is so low that the activity factor may be considered as equal to 1.

For $\ce{CaSO4}$, the effect is more pronounced, because of the presence of two doubly charged ions. The solubility calculated from the solubility product is 0.0051 M. The measured solubility, obtained from direct titration is 0.015 M. Three times more !

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  • $\begingroup$ How can we solve the above problem when we take into account the existence of intermediate ions? I appreciate your solution and calculation. $\endgroup$ – Adnan AL-Amleh Dec 10 '19 at 1:51
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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.

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    $\begingroup$ 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

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