Asked by Jessica
                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?
            
        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?
Answers
                    Answered by
            bobpursley
            
    http://en.wikipedia.org/wiki/Phosphate  You are looking at Ka2
You need to write the dissociation equation, and the expression for solubility constant.
    
You need to write the dissociation equation, and the expression for solubility constant.
                    Answered by
            Devron
            
    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^-]
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