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1. The problem statement, all variables and given/known data
Calculate the pH of 20.00 mL of 0.20 M KHP.
Given: pKa1(H2P)= 2.950 pKa2(H2P)= 5.408
2. Relevant equations
k = [H][P]/[HP]
3. The attempt at a solution
I used this equation and got:
3.91e-6 = [x][x]/[.2]
x^2 = 7.82e-7
x = 8.8e-4 = [H]
pH = 3.05
However, my homework is telling me this is incorrect. Is it just a rounding issue or is it something else? Any help is much appreciated.
http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt
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www.titrations.info (http://www.titrations.info) - all about titration (http://www.titrations.info) methods
symbolipoint
Nov5-09, 10:11 PM
Either your pK1 value is wrong OR your K1 value is wrong. Fix the wrong one and you may be able to improve your result. I base my comment on the idea that the first dissociation is far more important than the second dissociation in the solution just containing the dissolved KHP. I may be incorrect in this judgement (while still relearning).
This is Ka2 and pKa2. w330 tries to base calculations only on the second dissociation step, ignoring hydrolysis. As explained on the linked page this is incorrect approach.
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chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), concentration calculator (http://www.chembuddy.com/?left=CASC&right=concentration_and_solution_calculator)
www.titrations.info (http://www.titrations.info) - all about titration (http://www.titrations.info) methods
symbolipoint
Nov6-09, 01:52 PM
Borek, the discussion in your hyperlink in the post was interesting, and a little clearer than the discussion in the old analyitical textbook I read yesterday. Something almost like it was also shown in an old General Chemistry textbook. I have not seen much of that type of exercise for a long, long time.
In a way I am like an old analytical chemistry textbook :grumpy:
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