PH of 8.57x10-10M HNO3 solution?

  • Thread starter erjkism
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  • #1
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Homework Statement


I have to find the pH of a 8.57x10-10M HNO3 solution.

Homework Equations


HNO3 + H20 <--> H3O+ + NO3-
pH= -log[H3O+]

The Attempt at a Solution


my theory was that the hydronium ion concentration would be the hydronium ion concentration in water (1x10-14 M) plus the concentration of the nitric acid (8.57x10-10 M... because it fully dissociates). then take the negative log of it. i came up with a pH of 6.99

but my chemistry professor told me that i did it wrong and it is actually more complicated and could involve equilibrium. what did i do wrong?
 

Answers and Replies

  • #2
mgb_phys
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Are you sure you mean 10^-10 molar? that is a very low proportion (ie ppb) of acid
 
  • #3
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yeah thats what i mean. that's why this problem is getting difficult. if you just change the concentration of nitric acid to the H3O+ concentration, you get a pH above 7.. which is wrong
 
  • #4
symbolipoint
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yeah thats what i mean. that's why this problem is getting difficult. if you just change the concentration of nitric acid to the H3O+ concentration, you get a pH above 7.. which is wrong

Not so. Hydronium from the Nitric acid in water will give a pH BELOW 7.

You will need to use the dissociation of water along with the presence of the small concentration of the strong acid, nitric acid.
 
  • #5
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The water will not disassociate to produce 1x10^14 M H30+, it will disassociate to produce x M H30 and x M OH. You can then use the total molarity of H3O from the disassociation of water PLUS the disassociation of HNO3 multiplied the the molarity of OH to find Kw, from then you can solve the quadratic and find the pH.
 
  • #6
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pH = -log[H+] is a convention that only works if the concentration of the acid is 10^-6 or greater. Below 10^-6, the equation will never give you the correct answer. The case is the same with pOH.

For very dilute concentrations (greater than 10^-6), you need to consider equilibrium.
 
Last edited:

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