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JoannaBlackwelder
 one year ago
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.
JoannaBlackwelder
 one year ago
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.

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Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1This makes up a buffer right?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1\(\sf NaH_2PO_4 > Na^+ + H_2PO_4^\) \(\sf Na_2HPO_4 > 2Na^+ HPO_4_2^ \)

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1I 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 acidbase \( \sf H2PO4^ + H2O > HPO42^ + H3O^+\)

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1Na2HPO4 > 2Na^+ + HPO4^

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1We can use HendersonHasselbalch equation now for calculating the pH \(\sf pH = pK_a+log \frac{[conjugate~base]}{[acid]}\)

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1Ah, I see! That makes sense. :)

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1@Abhisar @JoannaBlackwelder What's the answer u get ?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1That's what I get. Why, what did you get?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Ummm.. [H2PO4] = 0.65/1.2 (acid) [HPO4] = 0.34/1.2 (conjugate base) I am getting around 7.05

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1Yeah I did it in the same way !

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1@JoannaBlackwelder is correct, pH = 6.9

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1We can use HendersonHasselbalch equation now for calculating the pH \(\sf pH = 7.21+log \frac{[0.28]}{[0.54]}\)

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1ooopss yes yes he is correct I have divided the acid moles by 12 !!!!!!!!

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1:) Thanks guys!

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1you get the same answer !!!! No I was wondering why my volume didn't cancel off . face palm
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