Finding amount of NaN02 needed to raise the a solution's pH to 5

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Discussion Overview

The discussion revolves around calculating the amount of sodium nitrite (NaNO2) needed to raise the pH of a 500 ml solution of 0.200 M nitrous acid (HN02) to a target pH of 5. The conversation includes theoretical considerations, equilibrium calculations, and the implications of adding a salt to a weak acid solution.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the required amount of sodium nitrite based on the desired pH and equilibrium constants but finds the result unexpectedly large.
  • Another participant suggests that the approach appears correct and notes that a significant increase in nitrate is expected due to the large change in pH from the pKa.
  • A participant corrects the initial mention of sodium nitrate to sodium nitrite and discusses the initial pH of the weak acid and the effect of adding sodium nitrite on the equilibrium.
  • One participant proposes that the equilibrium concentration of HN02 can be approximated as 0.2 M, arguing that a high concentration of NO2- will shift the equilibrium to the left.
  • Another participant questions whether the equilibrium concentration of HN02 would be significantly larger than its initial concentration due to the high excess of NO2-.
  • A response clarifies that there is no additional source of H+ ions other than water, which is a much weaker acid than HN02.

Areas of Agreement / Disagreement

Participants express differing views on the equilibrium concentration of HN02 after adding sodium nitrite, with some suggesting it remains close to 0.2 M while others propose it could be higher. The discussion does not reach a consensus on this point.

Contextual Notes

The calculations and assumptions made by participants depend on the specific conditions of the solution, including ionic strength and the behavior of weak acids in equilibrium. The implications of these factors on the final concentrations are not fully resolved.

rokku
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What amount of sodium Nitrate must be added to 500 ml of .200 M solution of HN02. Ka=4.0^-4


Ka=(H+)(A-)
--------
(HA)



Ph of 5 = 1.0^-5 (H+) so, (1.0^-5)(x)
---------- = 4.0^-4 , however X turns out to be way too big
.200
 
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Sorry, it is barely unreadable, try to use LaTeX to format the equation.

If I understand correctly what you did, your approach looks OK. Nothing strange in high amount of nitrate required - you want to get to pH which is 1.6 unit from the pKa, that requires around 101.6 more nitrate than there is nitrous acid.
 
Sorry about that, let me try again. Besides the obvious spacing issues I incorrectly mention Sodium Nitrate when it really should be Sodium Nitrite.

At first the pH of the undisturbed weak acid HN02 is around .69. I found it using a simple ICE table and the initial concentration HN02 (.200 Molar) plus the Ka (4.0^-4). When Sodium Nitrite is added, the salt completely dissociates and you end up with a common ion (NO2-) which would push the equilibrium to the left and result in less (H+) ions and a higher pH. Because we know what the desired pH is (5.00) we know that at equilibrium there should be the Antilog of 5, or 1.0^-5 M of H+. We also know that there has to be a new concentration of NO2- ions which is designated as X. Here is where I get stuck, what would the equilibrium concentration of HN02 be?

The formula to use would be Ka= (H+)(A-)/(HA)----> 4.0^-4=(1.0^-5)(X)/(equilibrium concentration of HN02?)
 
You can reasonably safely assume equilibrium concentration of HNO2 to be 0.2. High excess of NO2- shifts the dissociation far to the left.

My approximate calculations show that you need around 8.5M of NaNO2 (that ignoring ionic strength of the solution, which is pretty high).
 
Borek said:
You can reasonably safely assume equilibrium concentration of HNO2 to be 0.2. High excess of NO2- shifts the dissociation far to the left.

If High excess of NO2- shifts the dissociation far to the left, wouldn't the equilibrium concentration of HNO2 be significantly larger than its initial of .200?
 
No, you don't have a source of H+ (other than water, but it is an acid many orders of magnitude weaker than HNO2 itself).
 
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Ah I see, thank you.
 

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