🤓 Based on our data, we think this question is relevant for Professor Murray's class at UF.

We’re being asked to **calculate the K _{a}** of protonated urea (The pH of 1.00 M urea, a weak organic base is 7.050).

**Urea ((NH _{2})_{2}CO) is a weak base.**

Remember that * weak bases* partially dissociate in water and that

The dissociation of Urea is as follows:

(NH_{2})_{2}CO (aq) + H_{2}O(l) ⇌ (NH_{2})_{2}COH^{+} (aq) + OH^{-}(aq)

From this, we can construct an ICE table.

Remember that liquids are ignored in the ICE table.

**K**_{b}* expression* for (NH

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{\mathbf{}\mathbf{\left[}{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{2}}\mathbf{\right)}}_{\mathbf{2}}{\mathbf{COH}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{2}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{CO}\mathbf{\right]}}}$

*Liquids are ignored in the Kb expression.*

Note that each concentration is raised by the stoichiometric coefficient: [(NH_{2})_{2}CO], [OH^{–}] and [(NH_{2})_{2}COH^{+}] are raised to 1.

The pH of a 1.00 M solution of urea, a weak organic base, is 7.050.

Calculate the K_{a} of protonated urea.