So, it's a relatively common known solubility rule that any hydroxide with a cation not in the first two groups is basically insoluble. So supposing we have a transition or post-transition metal $M$ and say it forms a charge of $n+$ (so it's cation is $\ce{M^{n+}}$) why would $\ce{MX_n}$ (where $\ce{X}$ is a halide) remain in solution? Why doesn't the following reaction occur? $$ \ce{MX_n(aq)} + n\ \ce{H_2O(l)} \rightarrow \ce{M(OH)_n(s)} + n\ \ce{HX(aq)} $$ My thought was that perhaps this reaction does occur, but just very slowly and not fast enough for the $\ce{M(OH)_n}$ to precipitate in noticeable amounts. However, I don't really know and this is just a guess. Could someone explain to me why this doesn't happen?
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$\ce{MCl_n(aq)} + n\ \ce{H2O(l)} \rightarrow \ce{M(OH)n(s)} + n\ \ce{HCl(aq)}$ does tend to go to the right. "Basically insoluble" is not quantitative enough to generalize. If the hydroxide is very insoluble, the acid produced by hydrolysis may not keep it in solution. But if the hydroxide is even only slightly soluble, it may not precipitate out. The acidity of the solution is a subtle indicator that the reaction has shifted a little bit to the right. The reaction isn't slower, just not always so extreme.
If you put $\ce{SiCl4}$ in water, it gives $\ce{HCl}$ + silica gel because $\ce{SiO2}$ (or "$\ce{Si(OH)4}$") is so insoluble.
If you put $\ce{MgCl2}$ in water, no precipitate appears because $\ce{Mg(OH)2}$ is slightly soluble (~2 mg/L), and the $\ce{HCl}$ produced by the hydrolysis reaction is acidic enough so that if a significant amount of $\ce{Mg(OH)2}$ were produced, it would just redissolve. $\ce{Mg(OH)2}$ gives a pH of about 10 in water, so at pH's lower than 10, $\ce{Mg(OH)2}$ doesn't precipitate out. BTW, $\ce{MgCl2}$ is slightly acidic because of the hydrolysis reaction.
$\ce{AlCl3}$ is complicated because of ionic complexes, but in water it hydrolyzes to give a very acidic solution and some aluminum hydroxide complexes or gels which are of varying solubility. $\ce{AlCl3}$ has been used as a water treatment because it produces flocculent precipitates that carry down impurities.
$\ce{NiCl2}$ does not precipitate in water because $\ce{Ni(OH)2}$ is even more soluble than $\ce{Mg(OH)2}$.
$\ce{FeCl3}$ is quite acidic in water. You might expect the commercial 40% solution to be a cloudy product because of the production of acid, but it is a clear, deep red color. Oh, but when it is diluted and the pH rises because the acid is diluted, it precipitates in flocs like $\ce{AlCl3}$. It is used to clarify water by precipitating dirt and dust.
answered Feb 1, 2018 at 3:10
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$\begingroup$ How is nickel hydroxide more soluble than magnesium hydroxide? Ksp values show that at the same pH magnesium ions would be more soluble by about a factor of $10^4$. $\endgroup$ Feb 1, 2018 at 19:40
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$\begingroup$ I was surprised too. The CRC handbook gives 0.0009 g/100mL for Mg(OH)2; 0.013 g/100 mL for Ni(OH)2. I did not look up Ksp. I wonder if there is some other factor/difference in the measurements that makes sense. $\endgroup$ Feb 2, 2018 at 14:02
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$\begingroup$ If the numbers are right, nickel hydroxide in solution must be a weak base, while the limited amount of magnesium hydroxide that dissolves is much stronger. $\endgroup$ Feb 2, 2018 at 14:08