JoannaBlackwelder
  • JoannaBlackwelder
What is the pH of a solution made by the addition of 0.34 mole of Na2HPO4 and 0.65 mole of NaH2PO4 and sufficient water to give a total volume of 1.2 L? The pKa of H2PO4 is 7.21.
Chemistry
schrodinger
  • schrodinger
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Rushwr
  • Rushwr
This makes up a buffer right?
JoannaBlackwelder
  • JoannaBlackwelder
Yeah
Abhisar
  • Abhisar
\(\sf NaH_2PO_4 ----> Na^+ + H_2PO_4^-\) \(\sf Na_2HPO_4 ----> 2Na^+ HPO_4_2^- \)

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JoannaBlackwelder
  • JoannaBlackwelder
Right.
Abhisar
  • Abhisar
I am sorry, i don't know why the second latex is not working. pls bear with it.... Then we can see a case of bronsted acid-base \( \sf H2PO4^- + H2O --> HPO42^- + H3O^+\)
Rushwr
  • Rushwr
Na2HPO4 -----------> 2Na^+ + HPO4^-
Abhisar
  • Abhisar
We can use Henderson-Hasselbalch equation now for calculating the pH \(\sf pH = pK_a+log \frac{[conjugate~base]}{[acid]}\)
JoannaBlackwelder
  • JoannaBlackwelder
Ah, I see! That makes sense. :-)
JoannaBlackwelder
  • JoannaBlackwelder
Thank you!
Abhisar
  • Abhisar
Welcome c;
Rushwr
  • Rushwr
@Abhisar @JoannaBlackwelder What's the answer u get ?
JoannaBlackwelder
  • JoannaBlackwelder
6.93
Rushwr
  • Rushwr
are u sure ?
JoannaBlackwelder
  • JoannaBlackwelder
That's what I get. Why, what did you get?
Rushwr
  • Rushwr
I got 5.928
Abhisar
  • Abhisar
Ummm.. [H2PO4] = 0.65/1.2 (acid) [HPO4-] = 0.34/1.2 (conjugate base) I am getting around 7.05
Rushwr
  • Rushwr
Yeah I did it in the same way !
Abhisar
  • Abhisar
No, wait...
Abhisar
  • Abhisar
@JoannaBlackwelder is correct, pH = 6.9
Abhisar
  • Abhisar
We can use Henderson-Hasselbalch equation now for calculating the pH \(\sf pH = 7.21+log \frac{[0.28]}{[0.54]}\)
Rushwr
  • Rushwr
ooopss yes yes he is correct I have divided the acid moles by 12 !!!!!!!!
JoannaBlackwelder
  • JoannaBlackwelder
:-) Thanks guys!
Rushwr
  • Rushwr
you get the same answer !!!! No I was wondering why my volume didn't cancel off . face palm
Rushwr
  • Rushwr
Soo dumb

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