Posted by jean on Monday, January 30, 2012 at 4:34pm.
Hi I have a follow up to a question that was answered previously.
New question: So when the HPO4(2-) reacts with the NaOH, will the equation be this?:
HPO4^(2-) + NaOH --> H2O + NaPO4(2-)
But how do I write the equation for the H2PO4^- and HCl?
H2PO4^- + HCl -->
Here is the previous question and answer to refresh your memory.
write an equation for which component of the phosphate buffer (H2PO4-/ HPO4^2-) reacts when HCl is added. Do the same for NaOH.
I think that the H2PO4- will react with the NaOH but its only a guess and I'm confused as to how I would go about doing this.
chem - DrBob222, Sunday, January 29, 2012 at 5:21pm
You guessed wrong.
You have H2PO4^- and HPO4^2-
Which is the acid and which is the base? H2PO4^- is the acid. Why? Because it is the one with extra H^+. The HPO4^2- is the base because it is the one with fewer H (and the one accepting protons).
H2PO4^- ==> H^+ + HPO4^2-
This shows the H2PO4^- donating a proton and HPO4^2- accepting a proton.
- chem: Drbob222 - DrBob222, Monday, January 30, 2012 at 5:29pm
In a buffer made of H2PO4^- and HPO4^2-, addition of HCl adds to the base; i.e.,
H^+ + HPO4^2- ==> H2PO4^-
and addition of NaOH adds to the acid.
H2PO4^- + OH^- ==> H2O + HPO4^2-
I hope this gets things straight.
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