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$1.0\ \mathrm g$ of $\ce{AlCl3}$ is added to $100.0\ \mathrm{ml}$ of water. What would be the resulting pH?

I think the answer may be $4.022$ but I'm not certain.

The reaction wasn't provided but I think its this: $$\ce{Al(H2O6)6^3+ <=> H^+ + Al(H2O)5(OH)^2+}$$ Assuming all $\ce{Al}$ dissolved in water:
Initial $[\ce{Al}]=7.49\times10^{-4}\ \mathrm M$

$$x^2/(7.49\times10^{-4}-x)=1.4\times10^{-5}$$

$$x=9.56\times10^{-5}$$

$$\mathrm{pH}=-\log(9.56\times10^{-5})=4.022$$

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  • $\begingroup$ What happened to the Cl attached to the Al?!? $\endgroup$ – MaxW May 20 at 3:47
  • $\begingroup$ I have no idea, I'm so confused with this question $\endgroup$ – user93910 May 20 at 3:52
  • $\begingroup$ I think you have the right idea, but in an excess of water the overall reaction would be $$\ce{AlCl3 + 6H2O ->[aqueous] Al(H2O)5(OH)^{2+} + H+ + 3Cl-(aq)}$$ $\endgroup$ – MaxW May 20 at 4:00
  • $\begingroup$ Oh okay. Do you know how to solve this $\endgroup$ – user93910 May 20 at 4:03
  • $\begingroup$ Assume volume of solution is the same as the volume of water and that the reaction I gave goes to 100% completion. $\endgroup$ – MaxW May 20 at 4:07
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Not being given any other information assume the reaction goes to completion and that it is:

$$\ce{AlCl3 + 6H2O ->[excess H2O] Al(H2O)5(OH)^{2+} + H+ +3Cl-}$$

molecular mass $\ce{AlCl3 = \pu{133.34 g/mol}}$. So:

$$\ce{\pu{mol AlCl_{3}} = mol [Al(H2O)_5OH^{2+}] = mol [H+]} = \frac{\pu{1 g}}{\pu{133.34 g/mol}} = \pu{7.496\cdot 10^{-3} mol}$$

$$\pu{Molarity} = \dfrac{\pu{moles}}{\pu{volume}} = \dfrac{\pu{7.496\cdot 10^{-2} mol}}{\pu{0.100 L}} = \pu{0.07496 M}$$

pH = - log(0.07496) = 1.12


Now the truth is a bit more complicated. You can assume the reaction happens in two steps. The first step will go to completion.

$$\ce{AlCl3 + 6H2O ->[excess H2O] Al(H2O)6^{3+} +3Cl-}$$

The second step won't go to completion since $\ce{Al(H2O)6^{3+}}$ is a weak acid. There will be an equilibrium favoring the left side. But with no value for $K_\mathrm{eq}$ you are stymied.

$$\ce{Al(H2O)6^{3+} ->[excess H2O] Al(H2O)5(OH)^{2+} + H+}$$

$$K_\mathrm{eq} =\dfrac{\ce{[H+][Al(H2O)5(OH)^{2+}]}}{\ce{[Al(H2O)6^{3+}]}}$$

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  • $\begingroup$ oh ok? I'm very confused at this point $\endgroup$ – user93910 May 20 at 5:10
  • $\begingroup$ @nanxii - Sorry I fouled that up badly. $\endgroup$ – MaxW May 20 at 5:18
  • $\begingroup$ i'm kinda desperate at this point. I just need a simple explanation and answer $\endgroup$ – user93910 May 20 at 5:20
  • $\begingroup$ @nanxii - Since I messed up, I'll just give an answer. But this isn't a homework site. You've gotten a free ride on two questions. Don't expect more. See the site's homework policy. $\endgroup$ – MaxW May 20 at 6:06
  • $\begingroup$ Actually the OP has stated an equilibrium constant. The only problem i see is the miscalculation of the molarity. If we use the correct value you can calculate the pH assuming the first step goes to completition. (Wich is 3.0 considering the rounding for given significant digits.) $\endgroup$ – Andrew Kovács Jun 19 at 11:36

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