Predicting pH of an acidic salt of an weak acid and weak base
The main reactions which take place are

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

2)Hydrolysis of $\ce{HA-}$: $\ce{HA- + H+ -&gt; H2A}$

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

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 $k_{a1}&lt;&lt;[HCO_3^-]$ 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)