# Which of the following compounds should have the strongest conjugate acid? (See choices in answer).

Aug 4, 2015

The answer is indeed B. aniline.

#### Explanation:

The options are:

A. Ammonia ${K}_{b} = 1.8 \times {10}^{-} 5$
B. Aniline ${K}_{b} = 3.9 \times {10}^{-} 10$
C. Hydroxylamine ${K}_{b} = 1.1 \times {10}^{-} 8$
D. Ketamine ${K}_{b} = 3.0 \times {10}^{-} 7$
E. Piperidine ${K}_{b} = 1.3 \times {10}^{-} 3$

The strongest conjugate acid will correspond to the weakest base, which in your case is the base that has the smallest base dissociation constant, ${K}_{b}$.

For a generic weak base equilibrium, you have

${B}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s B {H}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

The base dissociation constant is defined as

${K}_{b} = \frac{\left[B {H}^{+}\right] \cdot \left[O {H}^{-}\right]}{\left[B\right]}$

The value of ${K}_{b}$ will tell you how willing a base is to accept a proton to form its conjugate acid, $B {H}^{+}$, and hydroxide ions, $O {H}^{-}$.

If more molecules of base ionize to form $B {H}^{+}$ and $O {H}^{-}$, then the numerator of ${K}_{b}$ will Increase. At the same time, the concentration of the base, which appears in the denominator, will decrease.

As a result, the value of ${K}_{b}$ will increase.

The stronger the base, the larger the concentrations of conjugate acid and hydroxide ions it produces in solution. An important consequence is that the equilibrium of stronger bases lies more to the right.

This implies that the reverse reaction

$B {H}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-} r i g h t \le f t h a r p \infty n s {B}_{\left(a q\right)}^{+} + {H}_{2} {O}_{\left(l\right)}$

is less likely to occur, meaning that the tendency of the base, $B$, to ionize will prevent too many $B {H}^{+}$ molecules to give up their proton to reform $B$ $\to$ $B {H}^{+}$ will be a weak conjugate acid.

On the other hand, if the base is weak, the equilibrium will lie more to the left. This means that the conjugate acid will be more willing to donate its proton to reform the base $\to$ $B {H}^{+}$ will be a strong conjugate acid.

So, the smaller the value of ${K}_{b}$, the smaller the degree of ionization of the base, meaning that base molecules tend to remain unionized in aqueous solution - not pick up a proton to form $B {H}^{+}$.

In your case, aniline will be weakest base, which implies that it will have the strongest conjugate acid.