# Problem with the solubility and Ksp of zinc hydroxide [closed]

My book gave me that solubility (g/L) is $$4.2\times 10^{-4}$$, so solubility (mol/L) is $$4.2\times10^{-6}$$. Then, using the chemical formula $$\ce{Zn(OH)2}$$, I know that the solubility product is $$K_\mathrm{sp} = \ce{[Zn^{2+}][OH-]^2}$$, so: $$K_\mathrm{sp} = (4.2\times10^{-6})(2\times(4.2\times10^{-6}))^2 = 3.0\times10^{-16}$$

However, the book says that the $$K_\mathrm{sp} = 3\times10^{-17}$$. Where did I go wrong?

• Direct relying of solubility and Ksp implies that 1) All Zn(aq) is as Zn^2+ 2) All OH- remains as OH- 3) Water autoionization plays no role 4) Data are for the same temperature and the best from the same source. If e.g there is present ZnOH+(aq), data cannot match well. – Poutnik May 17 at 9:47
• Also from Maurice's answer and comments (5) activity coefficients should be used instead of concentrations (6) ion pairing is negligible. – MaxW May 18 at 17:56
• – MaxW May 19 at 21:41

There are no simple relation between the solubility and the solubility product when a doubly charged ion is involved. The measured solubility is always much bigger than the value obtained from the solubility product. This is due to the doubly charged ion.

Doubly charged ions like $$\ce{Zn^{2+}}$$ are usually hydrolyzed in water, and are partly transformed into basic ions like $$\ce{[Zn(OH)]^+}$$ by the following reaction : $$\ce{Zn^{2+} + H2O -> [Zn(OH)]^+ + H+}$$ If $$\ce{Zn(OH)_2}$$ is dissolved in water, it will not only produce $$\ce{Zn^{2+}}$$ ions but also basic ions by the following equation : $$\ce{Zn(OH)_2 -> [Zn(OH)]^+ + OH^-}$$ As a consequence, the concentration of non complexed $$\ce{Zn^{2+}}$$ may be calculated from the solubility product of $$\ce{Zn(OH)_2}$$ but this $$\ce{[Zn^{2+}]}$$ is much smaller than the measured solubility of $$\ce{Zn(OH)_2}$$, because the measured solubility of $$\ce{Zn(OH)_2}$$ is the sum of $$\ce{[Zn^{2+}]}$$ and $$\ce{[Zn(OH)]^+}$$. It is even possible that ion pairs like $$\ce{Zn^{2+}(OH^-)_2}$$ are dissolved in water without being dissociated.

In the case of $$\ce{Zn(OH)_2}$$, the measured solubility is $$4.2 ·10^{-6}$$ M. And the concentration of $$\ce{Zn^{2+}}$$ taken from $$\ce{K_{sp} = 3·10^{-16}}$$ is $$\ce{[Zn^{2+}] = 1.95·10^{-6}}$$ M. This is much less than the measured solubility. It means that the total concentration of the dissolved ion pairs $$\ce{Zn^{2+}(OH^-)_2}$$ and of the basic ions $$\ce{Zn(OH)^+}$$ is equal to the difference $$(4.2 - 1.95)·10^{-6} = 2.25·10^{-6}$$ M. Only $$46.4$$% of all the dissolved $$\ce{Zn(OH)_2}$$ produce simple $$\ce{Zn^{2+}}$$ ions (which may be $$\ce{[Zn(H_2O)_6]^{2+}}$$ by the way).

• Magnesium and calcium hydroxides do seem to have solubilities close to the prediction for full ionization because both are sufficiently strong bases relative to their solubilities. But they are just about the only divalent metal hydroxides with both limited solubility and sufficient basic strength. – Oscar Lanzi May 17 at 14:08
• (1) Agree that overall notion is a possible explanation. (2) wouldn't say "There are no relation between the solubility and the solubility product when a doubly charged ion is involved." but rather the relationship can be complicated. (3) Not "doubly charged ion" but an ion that forms complexes. $\ce{Cu+}$ forms complexes, $\ce{Ca^{2+}}$ doesn't. (4) I'd expect $\ce{Zn(H2O)6^{2+}}$ in aqueous solution not "simple $\ce{Zn^{2+}}$ ions". – MaxW May 17 at 19:36
• oh ok, so my understanding is simplified :) – 12345bird May 18 at 5:48
• @MaxW. I agree with your remarks, except (3). I agree that $\ce{Cu^+}$ forms complexes. But the formation of a complex is not necessary to explain why the measured solubility is higher that the solubility calculated from $\ce{K_s}$. To take your example, namely Calcium, which does not form complexes, the solubility of $\ce{CaSO_4}$ is $4.7$ mM when calculated from $\ce{K_s}$ and $18$ mM when measured. Check : J. Chem. Educ. 7, 12, Dec. 2000, p. 1558. It is exactly like the case of $\ce{Zn(OH)_2}$ – Maurice May 18 at 9:37
• article link is doi.org/10.1021/ed077p1558.1 // Article discusses $\ce{CaSO4}$. // Not surprised that the activity coefficients have an effect. The ions parring is more influential due to a doubly charged cation and a doubly charged anion. I'd expect that ion pairing would affect $\ce{CaCl2}$ less. – MaxW May 18 at 17:49

The molar mass of zinc hydroxide is 99.424, so all your math checks out. Wikipedia also notes for zinc hydroxide that the $$K_\mathrm{sp} = 3.0\times 10^{-17}$$, so there is an inconsistency between the $$K_\mathrm{sp}$$ and the solubility data. Based on the $$K_\mathrm{sp}$$, the solubility should be $$\pu{1.9\times10^{-4} g/L}$$.

Obviously one source of the inconsistency is just different sources for the data. Another potential source of the is the temperature. It would take a careful literature search to find the original data sources and evaluate their relative quality to sort this out.

• ok thanks! so the two numbers dont "fit" together? – 12345bird May 17 at 7:52
• @12345bird - Impossible to say without a careful literature search. The Ksp could have been taken at $\pu{15 ^\circ C}$ and the solubility data at $\pu{25 ^\circ C}$. So in the absence of any other knowledge about the data all that can be said is that the values seem to be inconsistent. – MaxW May 17 at 9:15
• Another possibility is that $s = 2.1 \times 10^{-4}$ instead (half the original value). Thus, $K_\mathrm{sp} = 3.8 \times 10^{-17}$. – Mathew Mahindaratne May 17 at 9:33

@Maurice - Your last paragraph is inconsistent with $$K_\mathrm{sp} = 3.0\cdot10^{-17}$$. You gave:

$$\ce{[Zn^{2+}] = 1.95\cdot10^{-6}}$$ $$\ce{[Zn(OH)+] = 2.25\cdot10^{-6}}$$

Neglecting the autopyrolysis of water which is really an insignificant correction... $$\ce{[OH-] = (2\times1.95 + 2.25)\cdot10^{-6}} = 6.15\cdot10^{-6}$$ $$\therefore K_\mathrm{sp} = \ce{[Zn^{2+}][OH-]^2} = 7.4\cdot10^{-17}$$

Neglecting the the autopyrolysis of water means that the following approximation is used

$$\ce{[OH-] \approx 2\times [Zn^{2+}] + [Zn(OH)+]}$$

$$\ce{[OH-] = 2\times [Zn^{2+}] + [Zn(OH)+] + [H+]}$$

From solubility value $$\pu{4.2\cdot10^{−6} mol/L}$$, the $$K_\mathrm{sp} = 3.0\cdot10^{-17}$$, and including the autopyrolysis of water:

$$\ce{[Zn^{2+}] = 1.079\cdot10^{-6}}$$ $$\ce{[Zn(OH)+] = 3.121\cdot10^{-6}}$$ $$\ce{[OH-]} = 5.272\cdot10^{-6}$$

The pH would be 8.72 and only 26% of the zinc is $$\ce{Zn^{2+}}$$.

From the paper

the following image shows that three species ($$\ce{Zn^{2+}, Zn(OH)+, Zn(OH)2(aq)}$$) would have to be considered in this pH range. Also the activity coefficients and ion pairing would have some yet unknown effect. 