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PH of nitric oxide solution

  • Thread starter synergix
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1. Homework Statement

what is the pH of a 4.5x10-8M solution of nitric oxide?

3. The Attempt at a Solution

well first off..

NO(aq) + H2O(l) <-------> HNO+(aq) + OH-(aq)
or???........................-->HNO2(aq) + H+(aq)

Now I have no idea how much of the nitric oxide is actually going to react with the water. I could probably find a Ka value for this reaction online but that would be cheating there must be a way to figure this out without doing that.
 
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According to my notes. Nonmetallic oxides react with water to produce acids. according to Wikipedia this reaction occurs in water.
4 NO + O2 + 2 H2O ----> 4 HNO2
But the question doesn't say the NO is dissolved in water. I am lost
 

Borek

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Question is ambiguous. Even if NO is dissolved in water it is not clear whether there was oxygen present. So you can either assume it reacted to create equimolar amount of a weak acid, or assume it have not reacted with water at all and pH was not changed.

Which approach is correct depends on the teacher flexibility and sense of humor.
 
Well if I were to assume oxygen was present how would I calculate PH

if 2NO + O2 + H2O ---> 2HNO2

the Ka HNO2 is 4.5x10^-4

would the HNO2 then have to dissolve again back into solution so I could write a Ka equation for it?
 
so HNO2 + H2O ------> H3O+ + NO2
and then I can write my Ka expression and solve for [H3O+] ?
My teacher isn't incredibly flexible but I think it is reasonable to assume that some oxygen will be dissolved in water but it may usually be pretty insignificant. If I say this will only happen if sufficient oxygen is present he may still give me marks. it is a bonus question and we don't really do much on "oxides as acids or base."
 

Borek

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It doesn't have to dissolve back, you may assume reaction took place in the solution. Then it is just a solution of a weak acid.
 
How would I calculate the H3O + concentration without assuming it dissolves back and then writing the K a expression

Ka=4.5x10-4= [H3O+][NO2]/[HNO2]
 

Borek

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No idea what you mean by dissolves back. If it reacts in water, products are in water. It doesn't have to dissolve - it already is dissolved. It starts to dissociate after reaction and you will need Ka (and Kw) to find the solution pH.

Well... you will not need Ka. But you have to know why.
 
"No idea what you mean by dissolves back. If it reacts in water, products are in water. It doesn't have to dissolve - it already is dissolved."
Ya DUH! THX.
 
In this problem, since no specific data is given, I assume the standard conditions.

pH = -ln[H]

You already know the concentration (4.5x10-8 + 10-7)M
Just substitute and get the logarithm.:wink:
 

Borek

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You already know the concentration (4.5x10-8 + 10-7)M
You can't do that. Water autodissociation goes back in the presence of acids.
 

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