Phosphoric acid is triprotic acid (three ionizable hydrogens). The values of its stepwise ionization constants are Ka1= 7.5E-5, Ka2= 6.2E-8, and Ka3= 4.2E-13
1) Write the chemical equation for the first ionization reaction of phosphoric acid with water
2)Write the equilibrium constant expression (Ka1) for this reaction.
3)What would be the pH of a solution when [H3PO4]= [H2PO4-]?
(Note: pH= -log[H3O+])
5 answers
What is your trouble with this? What do you not understand?
I don't understand how to do the first one.
I really don't know how to do reactions
I really don't know how to do reactions
You know acid + water ionizes to give H3O^+ so take the H3O^+ out and see what's left.
H3PO4 + H2O ==> H3O^+ + H2PO4^-
Then Ka1 = (H^+)(H2PO4^-)/(H3PO4)
For the pH.
The pH when (H2PO4^-) = (H3PO4). Note one is in the numerator, the other is in the denominator, they cancel and you are left with Ka1 = (H^+), convert that to pH.
H3PO4 + H2O ==> H3O^+ + H2PO4^-
Then Ka1 = (H^+)(H2PO4^-)/(H3PO4)
For the pH.
The pH when (H2PO4^-) = (H3PO4). Note one is in the numerator, the other is in the denominator, they cancel and you are left with Ka1 = (H^+), convert that to pH.
thanx very much!!!!!
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3 answers