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Posted by on Tuesday, April 13, 2010 at 9:51pm.

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?

  • Chemistry - , Tuesday, April 13, 2010 at 10:34pm

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

  • Chemistry - , Tuesday, April 13, 2010 at 10:46pm

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

  • Chemistry - , Tuesday, April 13, 2010 at 11:03pm

    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.

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