http://en.wikipedia.org/wiki/Phosphate You are looking at Ka2
You need to write the dissociation equation, and the expression for solubility constant.
calculate oh- and ph in a solution in which dihydrogen phosphate ion is 0.335 M and hydrogen phosphate is 0.335 M.
This is a homework question, where I have the solution, which is 0.335/0.335, which I get, but then I am supposed to multiply this (1)by a number 6.2x10-8? I have absolutely no clue where this number is coming from. What is this number? How would I find it?
2 answers
I am not sure I understand what you are saying, but the number that number seems to be the Ka value for phosphoric acid.
pH=pka+log[A^-/HA]
let
pka=-log[Ka]
Ka=6.2 x 10^-8
pka=-log[Ka]=7.21
A^-=0.335
HA=0.335
To help you out a little bit:
The concentrations are equal, so the pka=pH
that is, the pH=7.21
pH+pOH=14
So, you can use that to solve for the pOH
pOH=-log[OH^-]
So, if you need the [OH^-] concentrations, use the above equation.
10^-(pOH)=[OH^-]
pH=pka+log[A^-/HA]
let
pka=-log[Ka]
Ka=6.2 x 10^-8
pka=-log[Ka]=7.21
A^-=0.335
HA=0.335
To help you out a little bit:
The concentrations are equal, so the pka=pH
that is, the pH=7.21
pH+pOH=14
So, you can use that to solve for the pOH
pOH=-log[OH^-]
So, if you need the [OH^-] concentrations, use the above equation.
10^-(pOH)=[OH^-]
1 answer