Weak acid-base solution assumption

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The discussion centers on calculating pH changes when mixing NaOH with HNO2, specifically using the equations for weak acid-base interactions. The user explores the impact of additional hydroxide ions on the equilibrium, leading to the need for simultaneous equations to accurately determine concentrations. The calculations reveal that the concentration of OH- from NaOH is 0.41345 M, while the interaction with H+ is significant enough to yield a 4.63% error if neglected. The conversation emphasizes the importance of rigorous approaches in acid-base equilibrium calculations.

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Conductivity
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I have been using a way to verify the assumptions.

For example let's say, If we add 200 mL of NaOH to 400 mL of HNO2 2M, The pH become 1.5 + (##pH_i##)
calculate the concentration of NaOH.

The usual way of doing is this:
## K_a = (0.02+x)[H^+]/(4/3-x) ## then find the x which most of the time is equal to the initial concentration of OH- after addition which you can convert back to the original solution and find the required information.

But the way I see how these approximations work is that you have to also think about how the additional OH can interact with H+ (or H3O+ if you like) to form water. So you really need simultaneous equations
##K_w = ([H+]_i -y)([OH-]_i-x-y) ##
Where X is from the original acid reaction and Y comes from forming water reaction.

Now usually Y is negligible so I can assume the OH- fully reacts with the acid but here in this situation Y isn't

I did the math and found that x = 0.41345 and y = 0.01917811 which is not negligible ( about 4.63% mistake, However I am talking about more extreme version of this) . Is this way of thinking is true?
 
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Conductivity said:
The pH become 1.5 + (##pH_i##)

Conductivity said:
## K_a = (0.02+x)[H^+]/(4/3-x) ##

No idea where these equations and numbers come from nor what the i means, care to elaborate?

In general when it comes to dealing with such calculations it is best to approach the system in a way described here: http://www.chembuddy.com/?left=pH-calculation&right=toc
 
Sure sorry,
First the concentration of NO2- in the acid solution alone is
## Ka * [HNO3]_i = x^2 ## solve for x, I get x = 0.03 (assuming [HNO3] doesn't change)

initial [NO2-] in the combined solution is 0.02 M
Initial [HNO3] let's assume it doesn't change so it is 4/3 M
Now when you add OH- the reaction will go to the right:
HNO2 (-----) H+ + NO2-
4/3-x ##[H+]_f## 0.02+x
Using the equilibrium constant equation we get the first equation.
After finding x which is 0.41345 M, It means that we had 0.41345 M of OH- from NaOH in the combined solution. Going back to the NaOH solution alone we get 0.620175

Note: Prepare for the mess! I know there are much simpler ways.

The other way that I am concerned about is this:
##[OH-]_i## represents the initial concentration of OH- that comes from NaOH in the combined solution
##[H+]_i## represents the initial concentration of H+ that comes from the disassociation of HNO2 in the combined solution before any reaction happen in the combined solution
If I form two simultaneous equation concerning that OH- might react with H+ or HNO2, I get
##\frac{K_a}{K_w} = \frac{(0.02+x)}{(4/3-x)([OH-]_i -x -y)}##
Where x results from HNO2 and OH- reaction

and Y results from this

##K_w = ([H+]_i -y)([OH-]_i-x-y) ##

Usually, Y is negligible but in this specific question it is not entirely negligible.
I found X = 0.41345 and Y = 0.019178
First, Is all this true?
and I am asking about if there was a extreme version of this, Where Y is not negligible at all. I have to do this in order to get an exact solution right.
 
Conductivity said:
Sure sorry,
First the concentration of NO2- in the acid solution alone is
## Ka * [HNO3]_i = x^2 ## solve for x, I get x = 0.03 (assuming [HNO3] doesn't change)

That is already based on an approximation, and you are mixing HNO3 with NO2- as if these were related - basically they are not.

I strongly suggest you read the linked pages before digging deeper, as it seems like you are just juggling equations, no wonder you are getting strange results.
 
Ah, I meant HNO2 with a ka of 4.8 * 10^(-4)
I made 2 parts one with an approximation and one without

Please see the second part. It must be true because I used it multiple times on very weak acid reaction and it gave exact answer. I want to know if this approach is true.

Yes I have seen some of topics you provided before. I also said in the beginning there is another way
Thanks in advance
 
You can't assume your[OH-]i (nor [H+]i) to be the only source of OH- (H+), once their concentrations go down acid (and base) will dissociate further (even if you ignore base dissociation here).

I told you to approach the problem in a more rigorous way, you are wasting time ATM.
 
I tried both ways and it gave the same answer.
So I guess both ways are true

I also found a way to simplify all what I wrote to 2 lines.
and I derived a equation to solve these kind of questions from your link

Thanks borek
 
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