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.

Question

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.

rushwr · Answer

This makes up a buffer right?

joannablackwelder · Answer

Yeah

abhisar · Answer

$\sf NaH_2PO_4 ----> Na^+ + H_2PO_4^-$
$\sf Na_2HPO_4 ----> 2Na^+  HPO_4_2^- $

joannablackwelder · Answer

Right.

abhisar · Answer

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 · Answer

Na2HPO4 -----------> 2Na^+ + HPO4^-

abhisar · Answer

We can use Henderson-Hasselbalch equation now for calculating the pH

$\sf pH = pK_a+log \frac{[conjugate~base]}{[acid]}$

joannablackwelder · Answer

Ah, I see! That makes sense. :-)

joannablackwelder · Answer

Thank you!

abhisar · Answer

Welcome c;

rushwr · Answer

@Abhisar  @JoannaBlackwelder  What's the answer u get ?

joannablackwelder · Answer

6.93

rushwr · Answer

are u sure ?

joannablackwelder · Answer

That's what I get. Why, what did you get?

rushwr · Answer

I got 5.928

abhisar · Answer

Ummm..
[H2PO4] = 0.65/1.2 (acid)

[HPO4-] = 0.34/1.2 (conjugate base)


I am getting around 7.05

rushwr · Answer

Yeah I did it in the same way !

abhisar · Answer

No, wait...

abhisar · Answer

@JoannaBlackwelder is correct, pH = 6.9

abhisar · Answer

We can use Henderson-Hasselbalch equation now for calculating the pH

$\sf pH = 7.21+log \frac{[0.28]}{[0.54]}$

rushwr · Answer

ooopss yes yes he is correct I have divided the acid moles by 12 !!!!!!!!

joannablackwelder · Answer

:-) Thanks guys!

rushwr · Answer

you get the same answer !!!! No I was wondering why my volume didn't cancel off . face palm

rushwr · Answer

Soo dumb