Calculating pH & pOH of AgNO2 Solution

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Discussion Overview

The discussion revolves around calculating the pH and pOH of a Silver Nitrite (AgNO2) solution when 0.1g is added to 100 mL of water. Participants explore the dissociation of NO2- ions and their impact on the acidity or basicity of the solution, considering the relevant equilibrium constants.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that NO2- ions will participate in an acid-base reaction with water, suggesting the solution will be basic.
  • Another participant questions the significance of the number of NO2- ions present and their effect on pH.
  • A participant proposes that despite the small Ksp, the equilibrium will shift to the right due to the presence of NO2- ions reacting with water, resulting in the formation of HNO2 and OH-.
  • Calculations are presented regarding the concentration of Ag+ and NO2- ions, leading to a proposed pOH and pH based on these values.
  • There is a correction regarding the initial concentration of NO2- ions, with one participant stating it should be .006499M instead of a previously calculated value.
  • Another participant verifies the calculation of Kb using the relationship Kb = Kw/Ka, expressing uncertainty about potential decimal place errors in previous calculations.
  • One participant suggests that the pH of the solution will be close to neutral, estimating it to be between 7 and 8.

Areas of Agreement / Disagreement

Participants express differing views on the expected pH and pOH values, with some calculations leading to significantly basic values while another participant suggests a near-neutral pH. The discussion remains unresolved with multiple competing views on the calculations and their implications.

Contextual Notes

Participants have not reached consensus on the correct approach to calculating pH and pOH, and there are unresolved questions regarding the assumptions made in the calculations, particularly concerning the solubility and dissociation of AgNO2.

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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


Homework Equations





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.
 
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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.
 

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