I was working through the end-of-chapter exercises of the acid-base chapter in Shriver's Inorganic Chemistry when I came across the following problem:
4.7. The effective proton affinity $\ce{A^{'}_{p}}$ of $\ce{F-}$ in water is 1150 kJ $\ce{mol^-}$. Predict whether it will behave as an acid or base in water.
I was initially stumped by this problem since the $\ce{A^{'}_{p}}$ of $\ce{F-}$ is between the $\ce{A^{'}_{p}}$ values of water and hydroxide ion (1130 kJ mol- and 1188 kJ mol-), which tempted me to say $\ce{F-}$ is amphoteric in neutral water even though I know it's basic from general chemistry.
The answer that the book's solution manual provides is the following:
The proton affinities of H2O (l) and OH-(aq) are 1130 kJ mol- and 1188 kJ mol-, respectively. Consequently, the reaction below between $\ce{H3O+}$ and $\ce{F-}$ is exothermic by -20 kJ mol- (1130 kJ mol- minus 1150 kJ mol-):
$\ce{ H3O+ (aq) + F- (aq) -> H2O (l) + HF (aq)}$
The reaction below between $\ce{H2O}$ and $\ce{F-}$ is endothermic by 38 kJ mol- (1188 kJ mol- minus 1150 kJ mol-):
$\ce{ H2O (l) + F- (aq) -> OH- (aq) + HF (aq)}$
Therefore, in neutral water, $\ce{F-}$ will behave as an base.
I understand that the first reaction has an approximate enthalpy change of -20 kJ mol- and is favored because it is the reverse reaction of the following reaction between $\ce{HF}$ and $\ce{H2O}$ and reaction between acids and $\ce{H2O}$ are favorable when the $\ce{A^{'}_{p}}$ of $\ce{H2O}$ is greater than the $\ce{A^{'}_{p}}$ of the conjugate base:
$\ce{ H2O (l) + HF (aq) -> H3O+ (aq) + F- (aq)}$
However, I have a few points I don't understand:
- Why does the text consider the second reaction with $\ce{H2O}$? Doesn't $\ce{F-}$ act as a base in both reactions?
- If both reactions occur and $\ce{F-}$ acts as a base, then which one is dominant?
- If the first reaction is dominant, then how does it proceed given the low concentrations ($\ce{1 x 10^{-7} M}$) of $\ce{H3O+}$ in neutral water?