# Calculate mass concentration of Ca(OH)2 solution

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?

• you have $0.0315$ mol of $\ce{Ca(OH)2}$ when it dissociates completely – Adnan AL-Amleh Aug 16 '18 at 21:33
• $m = 0.0315 \times 74.1$ to get 2.337 g – Adnan AL-Amleh Aug 16 '18 at 22:00
• As a previous comment said, calcium concentration is half of hydroxide concentration. It yields $0.0315$ then your procedure is correct! – santimirandarp Aug 17 '18 at 4:11
• Can someone please explain to me how the dissociation proceeds, so I understand how we arrive at the correct molar concentration – Bine Aug 18 '18 at 14:30
• 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. – 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.