# Why is magnesium sulfate acidic? Why are my calculations showing that it is basic?

Let's say I was dissolving $\pu{1M}~\ce{MgSO_4}$ into water and I wanted to find the pH. I would go right to the reaction of the dissociation of $\ce{Mg(OH)_2}$:

$$\ce{Mg(OH)_2 <=> Mg^2+ + 2OH-}$$

The $K_\mathrm{sp}$ expression for this reaction is:

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

According to wikipedia (see sources at bottom), the $K_\mathrm{sp}$ for $\ce{Mg(OH)_2}$ is $\pu{5.61*10^{-12}}$

$$\pu{5.61*10^{-12}}=[\ce{Mg^2+}][\ce{OH-}]^2$$

My molarity of $\ce{MgSO4}$ will be the same as that of $\ce{Mg^2+}$, which is $\pu{1 M}$. Substitute that in.

$$\pu{5.61*10^{-12}}=[\ce{OH-}]^2$$

And then solving for $[\ce{OH-}]$ we get:

$$[\ce{OH-}]=\pu{2.37*10^{-6} M}$$

Therefore:

$$\mathrm{pOH}=5.63$$ and $$\mathrm{pH}=8.37$$

So my final solution of $\ce{MgSO_4}$ will be basic. However, on stack exchange it says that the pH of $\ce{MgSO4}$ is between 5.5 and 6.5. The post does not explain why, but that the general consensus is that that is the range of the pH.

What did I do wrong?

Is magnesium sulfate basic, neutral or acidic?

• Simplified calculations which result in pH values between 6-8 (usually from extremely dilute acid/base solutions or by addtion of very weak acid/bases to water) are generally incorrect because they fail to take into consideration the autodissociation of water. Pure water starts at pH 7, and any hydroxide ions consumed by $\ce{Mg^2+}$ must cause the pH to fall below 7, as you would expect from adding an acid. Multiple previous answers here at Chem.SE touch on this point, for example here. You will have to find a slightly different equation. – Nicolau Saker Neto Jan 5 '18 at 5:08
• Using the Ksp of $\ce{Mg(OH)_2}$ would give you the maximum concentration of $\ce{OH-}$ for a solution of a given amount of $\ce{Mg^2+}$ - it won't necessarily give you the actual concentration. (It's perfectly happy with solubility products below the Ksp value.) -- If you're getting a pH above 7.0, you have to ask yourself where the "extra" $\ce{OH-}$ is coming from. It's from water, obviously, but what's driving the "extra" splitting of the $\ce{H2O}$, and - critically - what's happening to the $\ce{H+}$ that's being generated by the auto-dissociation of water to form the extra $\ce{OH-}$? – R.M. Jan 5 '18 at 19:32