# Calculating pH & pOH of AgNO2 Solution

• aqwsde

## Homework Statement

If 0.1g of Silver Nitrite (AgNO2; ksp = 6 x 10^-4) is added to 100 mL of water, what would the pH and pOH be? Note: The Ka for HNO2 is 7.2 x 10^-4

## The Attempt at a Solution

I really am not sure how to approach this except for the fact that and NO2- that dissociates will participate in a acid base reaction with water so it will be a basic solution.

Thanks.

So you know that NO2 is what will change the pH (and pOH). But what if we are only talking about a few NO2 ions? Will that change the pH by much? What if there are a whole bunch of NO2 ions available?

Try to determine how much N02 has dissolved and is floating around in the water ready to react with hydronium ions. Does the entire .1g of Silver Nitrate dissolve?

My thinking on the approach was that although the Ksp is small (i.e. not much will go into solution) since NO2- + H2O ----> HNO2 + OH- has a Kb = .1389 (Kw/Ka) it will shift the initial equilibrium of AgNO2 <---> Ag+ + NO2- to the right.

I still don't know where to go though and I really need to figure this out. I have my final tomorrow and really need to be able to understand this. Please help me out.

Ok Here is what I am thinking, let me know if this seems right:

.1gAgNO2/.1L*(1mol/153.875g)*(1mol Ag+/mol AgNO2) = .006499M Ag+ = x

Similarly .006499M NO2- = x

Q = x^2 = 4.2x10^-5 < Ksp => not saturated (all will disolve)

now [NO2-]i = 4.2x10^-5

setting up ice table yields:

Kb = .1389 = y^2/4.2x10^-5

y= 2.42x10^-3 = [OH-]
=> pOH = 2.62
pH = 11.38

now [NO2-]i = 4.2x10^-5
Why is this true?

woops [NO2-]i = .006499
y= .0300 = [OH-]
pOH = 1.52
pH = 12.35

Kb = .1389 (Kw/Ka)

Is this correct?
If Kb = kw / ka
Kb = 1x10^-14 / 7.2x10^-4
kb = 1.3889x10^-11

I wonder if you've missed a couple... decimal places :)

Do you have the answer for this question? I think I can help you but it's been awhile since I've done these kind of questions.

pH of the solution will be very close to neutral, between 7 and 8.