1
$\begingroup$

Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $\pu{10 mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

My idea was to calculate the amount of $\ce{Fe^3+}$ and $\ce{OH-}$ ions, and then add enough acid containing $\ce{H+}$ ions in order to neutralise the $\ce{OH-}$ ions corresponding to the amount of $\ce{Fe^3+}$. To do this I used the equilibrium constant leaving the hydroxide ions as the unknown. Then I calculated the difference between the initial hydroxide concentration (corresponding to $\pu{10 mg}$ of salt) and the needed concentration to reach $K_\mathrm{s}$. This corresponds to the concentration of hydronium ions in the acidic solution, and I can calculate pH.

Doing this I get $\mathrm{pH} = 3.03$. Is this how it should be done?

$\endgroup$
2
$\begingroup$

The OP first stated the problem as:

Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $\pu{10 mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

I took that to mean that acid was added to a solution containing 10 mg of iron (iii) hydroxide and the pH of the solution when the iron (iii) hydroxide was totally dissolved was desired.*

Knowing that $K_\mathrm{sp} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ then \begin{align} K_\mathrm{sp} &= \ce{[Fe^{3+}][OH^-]^3} = 6\cdot10^{-38},& \text{ so }&& \ce{[OH^-]} &=\sqrt[3]{\frac{6\cdot10^{-38}}{\ce{[Fe^{3+}]}}} \end{align}

You need to solve for $\ce{[Fe^{3+}]}$ which you can calculating knowing that there is $\pu{10 mg}$ of $\ce{Fe(OH)_3}$ in 0.1 liters of solution. It has a molecular weight of $\pu{107 g/mol}$, so there is $$0.010/107 = \pu{9.35* 10^-5 mol//L}\text{ of }\ce{Fe(OH)_3}.$$

Going back to the equation for the $K_\mathrm{sp}$: $$\ce{[OH^-]} =\sqrt[3]{\frac{6\cdot10^{-38}}{\ce{[Fe^{3+}]}}} = \sqrt[3]{\frac{6\cdot10^{-38}}{9.35\cdot10^{-5}}} = 8.63\cdot10^{-12}$$

Knowing $\ce{[OH^-]}$ you can calculate $\ce{[H^+]}$ via

$$\ce{[H^+]} = \frac{1\cdot10^{-14}}{\ce{[OH^-]}} = \frac{1\cdot10^{-14}}{8.63\cdot10^{-12}} = 1.16\cdot10^{-3}$$

and then pH:

$$\mathrm{pH} = -\log\left(1.16\cdot10^{-3}\right) = 2.94.$$


Now the OP has modified the problem statement to be:

Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $10\ \mathrm{mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

We again end up with the final pH being just acidic enough to dissolve $\pu{10 mg}$ of iron (iii) hydroxide in $\pu{100 ml}$ of water, but obviously the solution must start out more acidic. The final solution will again have a pH of $2.94$ or a $\ce{[H^+]} = 1.16\cdot10^{-3}$.

In $\pu{100 mL}$ of a solution of pH $2.94$ there are $\pu{0.116 mmol}$ of acid.

We calculated before that there were $\pu{9.35*10^{-5} mol}$ of $\ce{Fe^{3+}}$, but there are three $\ce{OH^-}$ anions for every cation of $\ce{Fe^{3+}}$, so there are $3 \times 9.35\cdot10^{-5} = \pu{2.81*10^{-4} mol}$ of $\ce{OH^-}$.

The total amount of substance (in millimoles) of acid needed is $$0.116 + 0.281 = 0.397.$$

For $\pu{100 ml}$ to contain $\pu{0.397 mmol}$ the solution would need to be 3.97 millimolar in $\ce{[H^+]}$, or the pH = $2.40$.

If you start with $\pu{100 ml}$ of solution with pH $2.40$ and dump in $\pu{10 mg}$ of $\ce{Fe(OH)3}$, then the final pH will be $2.94$ and all of the $\ce{Fe(OH)3}$ will be just dissolved.

$\endgroup$
  • $\begingroup$ Thank you for your answer. I understand what the steps are, I have done more exercises like this. But what I still don't really have 100% clear is how you can calculate the concentration of Iron (III) ions without involving the constant again, as you start off with the actual $\ce{Fe(OH)3}$. $\endgroup$ – Bee Jan 15 '17 at 8:25
  • $\begingroup$ HINT - Fe has an atomic weight of 55.85 and $\ce{Fe(OH)_3}$ has a molecular weight of 106.9, so 10 mg of $\ce{Fe(OH)_3}$ has (55.85/106.9)*10 = 5.22 mg of Fe. $\endgroup$ – MaxW Jan 15 '17 at 8:53
  • $\begingroup$ Okay, I think I got it. Just to be clear: whenever I have solubility equilibrium problems like this one, should I always try to find the value of the concentrations that make Q = K? Because if I am not mistaken, initially, for this exercise, Q > K. And according to my theory notes, when this happens, precipitation, not dissolution, will take place until Q = K. $\endgroup$ – Bee Jan 15 '17 at 9:41
  • $\begingroup$ Q=K is the knife's edge on these types of problems. So you'd want pH to be "less than" whatever is calculated. $\endgroup$ – MaxW Jan 15 '17 at 15:54
  • $\begingroup$ Just something I noticed yesterday: in the equilibrium expression, why don't you also multiply by 27, as the concentration of hydroxide ions is three times that of iron ions? And also, why do you calculate the hydroxide ion concentrations via the water constant expression? As you are trying to neutralise, shouldn't it be the same concentration as hydroxide ions? $\endgroup$ – Bee Jan 17 '17 at 9:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.