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

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To raise the pH of a 500 ml solution of 0.200 M HNO2 to 5.00, a significant amount of sodium nitrite (NaNO2) is required due to the common ion effect, which shifts the equilibrium left and reduces H+ concentration. The initial pH of the undisturbed HNO2 solution is approximately 0.69, and the desired concentration of H+ at pH 5.00 is 1.0 x 10^-5 M. Calculations indicate that around 8.5 M of NaNO2 is needed, although this does not account for the solution's ionic strength. The equilibrium concentration of HNO2 can be assumed to remain close to 0.200 M, as the high excess of NO2- ions significantly influences the dissociation. Understanding these dynamics is crucial for accurately adjusting the pH of the solution.
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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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