Question
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?
Answers
DrBob222
The pH of a NaHCO3 solution is independent of the concn. pH = 1/2(pk1 + pk2)
HELP
is this because this acts as a polyprotic acid, or because the ratio in the Henderson Hasselbach for HA/A- remains the same?
DrBob222
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.
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.