Calculate the pH of 20.00 mL of 0.20 M KHP

[w3390]

Homework Statement

Calculate the pH of 20.00 mL of 0.20 M KHP.

Given: pKa1(H2P)= 2.950 pKa2(H2P)= 5.408

Homework Equations

k = [H][P]/[HP]

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.

 

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

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methods

 

[symbolipoint]

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).

 

[Borek]

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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methods

 

[symbolipoint]

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.

 

[Borek]

In a way I am like an old analytical chemistry textbook :grumpy: