Answer : The pH will be, 3.2
Explanation :
As we known that the value of solubility constant of ferric hydroxide at [tex]25^oC[/tex] is, [tex]2.79\times 10^{-39}[/tex]
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.
[tex]1.5mg/L=\frac{1.5\times 10^{-3}g/L}{56g/mol}=2.7\times 10^{-7}mol/L[/tex]
The chemical reaction will be:
[tex]Fe(OH)_3\rightarrow Fe^{3+}+3OH^-[/tex]
The expression of solubility constant will be:
[tex]K_{sp}=[Fe^{3+}]\times [3OH^-]^3[/tex]
Now put all the given values in this expression, we get the concentration of hydroxide ion.
[tex]2.79\times 10^{-39}=(2.7\times 10^{-7})\times [3OH^-]^3[/tex]
[tex][OH^-]=1.5\times 10^{-11}M[/tex]
Now we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (1.5\times 10^{-11})[/tex]
[tex]pOH=10.8[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-10.8\\\\pH=3.2[/tex]
Therefore, the pH will be, 3.2
