[Anthasci]

Greetings! So, I'm trying to understand a little thing here:

100 mL of NaH₂BO₃ (0,100 M) is poured over 5,0 g of a strong cation exchanger (i.e.), with the capacity of 40,0 mg CaO/g. What is the pH of the eluate after we've added 10,0 mL of 0,100 M H₃BO₃ to it?

My train of thought went like this;

calculating the total capacity of the i.e. yields 3,565 mmol of CaO, equalling 7,130 mmol for Na⁺. Given that we have 10,0 mmol of NaH₂BO₃, the eluate should contain:
-2,87 mmol of H₂BO₃^- (the salt that didn't exchange it's Na+)
-7,13 mmol of H₃BO₃ (equal to the amount in moles of Na+ that exchanged with the i.e., thus protonating the salt)

and since pKa₁ = 9,24

the pH of this eluate is 8.84

now when we add the acid, does the Henderson-Hasselbalch equation equal:

pH = pKa₁ + log [itex] \frac {[H₂BO₃^-] - [H₃BO₃]_added}{[H₃BO₃] + [H₃BO₃]_added}[/itex]

of course taking into account that we added 1 mmol of H₃BO₃ and that the total volume is now V=0,110 L, and we arrive at:

pH=8.60

My main dilemma here is as follows: When we have an acid/conjugated base buffer and we add strong acid, in the numerator, we subtract the amount of protons added and add this amount in the denominator. From my understanding, when I'm adding that extra H₃BO₃ I'm adding HA, which is like adding protons, is this correct? Can I just add this number in the denominator and subtract in the numerator? And would the reverse hold if I were to be adding the salt of H₂BO₃^-?

Thank you very much for tips and help

[Borek]

You are not protonating H₂BO₃^- adding H₃BO₃. Even if you would, you will produce exactly the same substances in exactly the same amounts (so the net effect is null). The only thing that you are changing by addition of H₃BO₃ is the amount of the H₃BO₃ present.

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