equilibrium - Predicting pH of an acidic salt of an weak acid and weak base

Suppose we want to calculate pH of an $\ce{NH4HCO3}$ solution of known concentration. Now, $\ce{NH4+}$ will be hydrolised to give $\ce{H+}$ ions. Some $\ce{HCO3-}$ ions will be converted to (i) $\ce{H2CO3}$ and $\ce{OH-}$ and some will give (ii) $\ce{H+}$ and $\ce{CO3^2-}$

\begin{align} \ce{HCO3- + H2O &<=> H2CO3 + OH− }& 1/K_\mathrm{a1}\\ \ce{HCO3- + H2O &<=> CO3^2- + H3O+ }& K_\mathrm{a2}\\ \ce{NH4+ + H2O &<=> NH3 (aq) + H3O+ }& 1/K_\mathrm{b}\\ \end{align}

1. 
   

   I am not sure how to take account of both $\ce{H3O+}$ and $\ce{OH-}$ in the calculation of pH.

   
   


2. 
   

   Will the third equilibium costant be $1/K_\mathrm{b}$?

   
   

Answer

The main reactions which take place are

1)Ionization of $\ce{HA-}$: $\ce{HA- -> H+ +A^2-}$

2)Hydrolysis of $\ce{HA-}$: $\ce{HA- + H+ -> H2A}$

3)Hydrolysis of cation: $\ce{B+ + H2O -> BOH + H+}$

for reaction $1$, $$k_{a2}=\frac{[H^+][A^{2-}]}{[HA^-]}$$

for reaction $2$, $$k_{a1}=\frac{[HA^-][H^+]}{[H_2A]}$$

for reaction $3$, $$k_h=\frac{k_w}{k_b}=\frac{[BOH][H^+]}{[B^+]}$$

At any moment, $$\ce{[H^+]=[BOH] +[A^{2-}]-[H2A]}$$ $$[H^+]=\frac{[NH^{4+}].k_w}{[H^+]k_b}+\frac{k_{a2}.[HCO_3^-]}{[H^+]}-\frac{[HCO_3^-][H^+]}{k_{a1}}$$ Rearranging the terms to get $\ce{[H+]}$, $$[H^+]=\sqrt{\frac{\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]}{1+\frac{[HCO_3^-]}{k_{a1}}}}$$ $$[H^+]=\sqrt{\frac{k_{a1}\cdot \big(\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]\big)}{k_{a1}+[HCO_3^-]}}$$

Now we will have to make 2 assumptions. First lets assume that $k_{a1}<<[HCO_3^-]$ so that $k_{a1}+[HCO_3^-]=[HCO_3^-]$

$$[H^+]=\sqrt{\frac{k_{a1}\cdot \big(\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]\big)}{[HCO_3^-]}}$$

If you assume that neither hydrolysis nor the dissociation goes too far, you can assume the concentration of bicarbonate ion to be equal to the initial concentration (say C). Since Ammonium bicarbonate dissociates to give two products in 1:1 ratio, concentration of ammonium produced is equal to the concentration of bicarbonate ion produced.

I saw this formula for calculation pH of such salts in my textbook but no derivation was given: $$[H^+]=\sqrt{k_{a1}\cdot\big(\frac{k_w}{k_b}+k_{a2}\big)}$$ The above formula can be obtained if you consider the last assumption. (Clarification needed)