Question

Equilibrium constant for the acid ionization of ${Fe}^{3+}$ to $Fe{\left(OH\right)}^{+2}$ and $H^{+}$ is $6.5\times {10}^{-3}$. What is the maximum $pH$ which could be used so that at least $95$% of the total ${Fe}^{3+}$ in a dilute solution exists as ${Fe}^{3+}$?

• A. $pH=0.908$
• B. $pH=4.75$
• C. $pH=9.25$
• D. $pH=13.1$

Answer 1

Answer: (A)

$pH=0.908$

As given, ${Fe}^{+3}+H_{2}O⟶Fe{\left(OH\right)}^{+2}+H^{+}$ ; $K=6.5\times {10}^{-3}$

$0.95x$ $0.05x$ $0.05x$
(If dissociation is x)

Now

$K=\frac{\left[Fe\right(OH{)}_2^{+2}\left]\right[H^{+}]}{\left[F{e}^{+3}\right]}$

$6.5\times {10}^{-3}=\frac{{\left(0.5\right)}^{2}x}{0.95}$

$x=2.47$

$\left[H^{+}\right]=0.05x=0.1235$

Using $pH=-log\left[H^{+}\right]=-log\left(0.1235\right)$

$pH=0.908$