Question

Groundwater in Pherric, New Mexico, initially contains 1.800 mg/L of iron as Fe3+. What must the pH be raised to in order to precipitate all but 0.3 mg/L of the iron? The temperature of the solution is 25˚C.

Answer 1

Answer : The pH will be, 3.2

Explanation :

As we known that the value of solubility constant of ferric hydroxide at $25^oC$ is, $2.79\times 10^{-39}$

Amount or solubility of iron consumed = (1.800 - 0.3) mg/L = 1.5 mg/L

The given solubility of iron convert from mg/L to mol/L.

$1.5mg/L=\frac{1.5\times 10^{-3}g/L}{56g/mol}=2.7\times 10^{-7}mol/L$

The chemical reaction will be:

$Fe(OH)_3\rightarrow Fe^{3+}+3OH^-$

The expression of solubility constant will be:

$K_{sp}=[Fe^{3+}]\times [3OH^-]^3$

Now put all the given values in this expression, we get the concentration of hydroxide ion.

$2.79\times 10^{-39}=(2.7\times 10^{-7})\times [3OH^-]^3$

$[OH^-]=1.5\times 10^{-11}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1.5\times 10^{-11})$

$pOH=10.8$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-10.8\\\\pH=3.2$

Therefore, the pH will be, 3.2