Calculate the amount of $\ce{Mg(OH)2}$ which is solubilized in $\pu{1.25 L}$ of a buffered solution at $\mathrm{pH} = 9.5$. $(K_\mathrm{sp}(\ce{Mg(OH)2}) = \pu{5.61E-12})$.

From $\mathrm{pH}$ I compute the $\mathrm{pOH}$, so $[\ce{OH-}] = \pu{3.16E-5}$. But how can I compute $[\ce{Mg(OH)2}]$?

$[\ce{Mg}^{2+}]=\frac{K_\mathrm{sp}}{[\ce{OH-}]^2}$ and then?


  • 2
    $\begingroup$ You compute [$\ce{Mg^2+}$] not [$\ce{Mg(OH)2}$] // You may find useful these links for text formatting (not to be applied to titles): Notation basics , Formatting of math/chem expressions and upright vs italic $\endgroup$
    – Poutnik
    Jan 12, 2022 at 14:00
  • $\begingroup$ I need $[\ce{Mg(OH)2}]$ for the solution, no? $\endgroup$
    – Rick88
    Jan 12, 2022 at 14:34
  • $\begingroup$ If [$\ce{Mg^2+}$] is e.g. $\pu{1 mmol L-1}$, then $\pu{1 mmol}$ of $\ce{Mg(OH)2(s)}$ has been dissolved in 1 L of the buffer. $\endgroup$
    – Poutnik
    Jan 12, 2022 at 14:38
  • $\begingroup$ Thus $[\ce{Mg^2+}]=\frac{[\ce{OH-}]}{2}$, then $[\ce{Mg^2+}]=[\ce{Mg(OH)2}]$? $\endgroup$
    – Rick88
    Jan 12, 2022 at 15:00
  • $\begingroup$ Neither, neither. $\endgroup$
    – Poutnik
    Jan 12, 2022 at 15:01


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