How would the pH differ for a 0.1M solution of NaHCO3 vs 0.2M NaHCO3 - remain the same, double, half,or can't be calculated without Ka or Kb?

The pH of a NaHCO3 solution is independent of the concn. pH = 1/2(pk1 + pk2)

is this because this acts as a polyprotic acid, or because the ratio in the Henderson Hasselbach for HA/A- remains the same?

It does act as a polyprotic acid (really diprotic) and I suppose one could argue that HA = A^- but I don't think that will hold. What you do is write down k1 and k2, then multiply k1*k2. If you will do that it ends up like this.

k1k2 = (H^+)(HCO3^-)/(H2CO3)x(H^+)(CO3^-2)/(HCO3^-).

So (H^+)*(H^+)=(H^+)^2, then (HCO3^-) in numerator and denominator cancels. It turns out that (CO3^-2) left in the numerator = (H2CO3) in the denominator and we are left with
k1k2 = (H^+)^2.
With the salts of H3PO4, it works the same way.
NaH2PO4 in water is sqrt(k1k2) while Na2HPO4) in water is sqrt(k2k3). You can go through similar derivations for those.

To determine how the pH would differ for the given solutions, we first need to understand the properties of NaHCO3 (sodium bicarbonate).

NaHCO3 is a salt that consists of a weak acid (H2CO3, carbonic acid) and its conjugate base (HCO3-, bicarbonate ion). When dissolved in water, NaHCO3 undergoes hydrolysis, meaning it reacts with water and produces OH- (hydroxide ions) or H3O+ (hydronium ions). The pH of a solution depends on the concentration of H3O+ ions present.

To calculate the pH, we need to consider the dissociation of NaHCO3 into HCO3- and Na+ ions. The reaction can be represented as follows:

NaHCO3 ⇌ HCO3- + Na+

At equilibrium, a fraction of NaHCO3 dissociates, resulting in the formation of HCO3- ions. Since NaHCO3 is a weak acid, it only partially dissociates in water. The degree of dissociation depends on the concentration of the solution.

Now let's consider the provided solutions:

1) 0.1 M NaHCO3 solution:
In this case, we have a lower concentration of NaHCO3. With a lower concentration, less NaHCO3 will dissociate into HCO3- ions, resulting in a lower concentration of H3O+ ions. Consequently, the pH of the solution will be higher (more basic).

2) 0.2 M NaHCO3 solution:
Here, the concentration of NaHCO3 is double that of the previous solution. With a higher concentration, more NaHCO3 will dissociate into HCO3- ions, leading to a higher concentration of H3O+ ions. Therefore, the pH of the solution will be lower (more acidic).

To summarize, the pH of a 0.1 M NaHCO3 solution would be higher (more basic), while the pH of a 0.2 M NaHCO3 solution would be lower (more acidic).