Suppose we have a strong basic aqueous solution $B$ that we attempt to neutralize with $\ce{NaHCO3}$ working as an acid $A$.
The source of excess $\ce{OH-}$ ions comes from the base, and since it's strong, we can consider:
$$n_{Bo}=n_{\ce{OH-}}\implies C_{Bo}{V_B}=C_{\ce{OH-}}V_B$$
The source of excess $\ce{H+}$ ions is bicarbonate, but since it's a weak acid, we have to consider the equilibrium:
$$\ce{HCO3-(aq)<=>H+(aq) + CO3^{2-}(aq)}$$
The equilibrium expression in terms of moles after mixing both solutions is:
$$K_n=K_a\;(V_A+V_B)=\frac{x^2}{C_{Ao}V_A-x}$$
For complete neutralization to take place, both excess $\ce{H+}$ and excess $\ce{OH-}$ ions need to completely react with each other, so:
$$x=C_{Bo}V_B$$
Substituting above:
$$K_a\;(V_A+V_B)=\frac{(C_{Bo}V_B)^2}{C_{Ao}V_A-C_{Bo}V_B}$$
For simplicity, let's consider equal molar concentrations of both solutions and a volume of $\pu{0.1L}$ for our basic solution that needs neutralizing:
$$C_{Ao}=C_{Bo}=\pu{1mol/L}$$
$$V_B=\pu{0.1L}$$
The acid dissociation constant of bicarbonate at 25°C is approximately:
$$K_a=5.012\times10^{-11}$$
Substituting all values:
$$5.012\times10^{-11}\;(V_A+0.1)=\frac{0.1^2}{V_A-0.1}$$
Solving for $V_A$:
$$V_A=\pu{14125.2L}$$
In other words, bicarbonate is considerably weaker as an acid (about 400 times) than it is as a base, so attempting to neutralize a strong basic solution with it would require an enormous amount.