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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! $\endgroup$ – 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
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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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