# Finding dissolved amount of base from pH and solubility product

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?

$$[\ce{Mg(OH)2}]=[\ce{Mg^2+}]+[\ce{OH-}]^2$$?

• 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 Commented Jan 12, 2022 at 14:00
• I need $[\ce{Mg(OH)2}]$ for the solution, no? Commented Jan 12, 2022 at 14:34
• 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. Commented Jan 12, 2022 at 14:38
• Thus $[\ce{Mg^2+}]=\frac{[\ce{OH-}]}{2}$, then $[\ce{Mg^2+}]=[\ce{Mg(OH)2}]$? Commented Jan 12, 2022 at 15:00
• Neither, neither. Commented Jan 12, 2022 at 15:01