A question in my college test was whether $\ce{MgSO4}$ is neutral, acidic or basic. I was told to solve this kind of problem by hydrolyzing any ions that do not "come from" a strong acid or a strong base, and if nothing is hydrolyzed, then the salt is neutral. So by this logic, as $\ce{Mg^2+}$ comes from $\ce{Mg(OH)2}$ and $\ce{SO4^2-}$ comes from $\ce{H2SO4}$, I answered "neutral".

My professor marked it as wrong, and said that the salt $\ce{MgSO4}$ is basic. How can you explain this?

  • 2
    $\begingroup$ Hint: weak bases are usually insoluble. $\endgroup$
    – permeakra
    Apr 20, 2016 at 8:13
  • 3
    $\begingroup$ Basic makes no sense at all. The only way for the solution to be basic is for $\ce{SO4^{2-}}$ to hydrolyze to $\ce{HSO4^{-}}$. The pKa2 for sulfuric acid is 1.99, so that isn't going to happen in a solution of $\ce{SO4^{2-}}$. $\endgroup$
    – MaxW
    Mar 1, 2017 at 19:48
  • $\begingroup$ @MaxW Suppose you start with neutral water, so that initially hydroxide and hydronium ions are equal. You then add the sulfate anion, the equilibrium constant for the following equation is $K_b = K_w/K_a$ so that $pKb = 12.01 $: $ \\ SO4^{2-} + H2O \implies HSO4^{-} + OH^{-}$ Although you are correct that the reverse direction here is favored, bisulfate is a weak enough acid for the forward direction to be non-negligible. In the process of reaching equilibrium, there will be a generation of hydroxide ions. Since the solution was initially neutral, the final solution is slightly basic.... $\endgroup$
    – David Reed
    Nov 13, 2019 at 1:10
  • 1
    $\begingroup$ … This analysis ignores the contribution of the cation. In this instance the two oppose each other, Mg works to make it acidic whereas sulfate attempts to make it basic. Then you have to do a computation to see which one "wins". Had it been sodium sulfate the answer WOULD be basic because sodium hydroxide disassociates completely beneath its saturation limit and thus it would be analogous to a "strong acid" in that the reverse direction in which a sodium ion bonds with a hydroxide ion to form sodium hydroxide is negligible and can be fairly said not to occur. $\endgroup$
    – David Reed
    Nov 13, 2019 at 1:19

3 Answers 3


Looking at the following references, it seems that most agree that a water solution of $\ce{MgSO4}$ is slightly acidic.

"The pH of an aqueous magnesium sulfate solution is related to the molarity of the MgSO4. Typically, the pH is between 5.5 and 6.5 due to magnesium's affinity for hydroxide ion (OH-). As the sulfate goes into solution, hydroxide anions associate with the magnesium, increasing the relative ratio of H+ to OH-. This shift results in more acidic solutions."

"The pH of a pure magnesium sulfate solution is approximately 6.4"

"The pH of [magnesium sulfate] hydrates is average 6.0 (5.5 to 6.5)."

"pH: Neutral to litmus"

Why not perform the experiment in class, using an accurate pH meter and/or indicators (see the pH indicator chart in Wikipedia for suggestions).

  • $\begingroup$ +1 for nice answer and all the references. // I'm not so sure that the pH meter experiment would be illustrative. A very low amount of impurities could shift pH acidic or basic. So how do you obtain salt of "sufficient" purity? $\endgroup$
    – MaxW
    Mar 2, 2017 at 1:00
  • $\begingroup$ "The pH of a pure magnesium sulfate solution is approximately 6.4, a pH distinctly more acidic being obtained by adding e.g. 1 wt% FeSO4 • H2O (pH: 4.6) or 1 wt% KH 2PO 4 (pH: 3.2)." There is no evidence given for the pH of about 6.4, and they seem to think that adding $\ce{KH2PO4}$ makes the pH more acidic perhaps through some interaction, which is weird. $\endgroup$ Mar 20 at 16:25
  • $\begingroup$ The wikipedia article no longer makes a claim for the pH of dissolved hydrates. $\endgroup$ Mar 20 at 16:27
  • $\begingroup$ Pubchem: "Efflorescent crystals or powder; bitter, saline, cooling taste; density: 1.67; pH 6-7; soluble in water (g/100 ml): 71 @ 20 °C, 91 @ 40 °C; slightly soluble in alcohol; its aqueous soln is neutral; ". So they say the pH is between 6 and 7, but then say that the aqueous solution is neutral. $\endgroup$ Mar 20 at 16:31

Aqueous solution of magnesium sulfate is going to be slightly acidic, and there is no need to even go to the lab to test it. $\ce{MgSO4}$ is a salt formed by a weak base and a strong acid (both dibasic), so there is a two-step hydrolysis of magnesium cation.

First and primary step is the formation of a basic salt:

$$ \begin{align} \ce{2 MgSO4 + 2 H2O &<=> (MgOH)2SO4 + H2SO4} \tag{R1.1}\\ \ce{2 Mg^2+ + 2 SO4^2- + 2 H2O &<=> 2 MgOH+ + SO4^2- + 2 H+ + SO4^2-}\tag{R1.2}\\ \ce{Mg^2+ + H2O &<=> MgOH+ + H+}\tag{R1.3} \end{align} $$

Second minor (negligible) step:

$$ \begin{align} \ce{(MgOH)2SO4 + 2 H2O &<<=> 2 Mg(OH)2 + H2SO4}\tag{R2.1}\\ \ce{2 MgOH + SO4^2- + 2 H2O &<<=> 2 Mg(OH)2 + 2 H+ + SO4^2-}\tag{R2.2}\\ \ce{MgOH+ + H2O &<<=> Mg(OH)2 + H+}\tag{R2.3} \end{align} $$

Here indices .1, .2, .3 denote molecular, complete ionic, and net ionic equation, correspondingly. As a result of the hydrolysis reaction, some excess amount of hydronium cations appears, therefore, the the medium is expected to be acidic.


$\ce{SO4^2-}$ is actually going to make the pH basic because $\ce{HSO4-}$ is a weak acid (not $\ce{H2SO4}$, only the first hydrogen ionizes completely). By itself aqueous $\ce{Mg^2+}$ will not form magnesium hydroxide in water. But with the $\ce{OH-}$ from the $\ce{HSO4-}$ is will form a suspension with some dissolved and some not so dissolved, but still the solution is basic.


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