Titration of Nitrous Acid with Sodium Hydroxide

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In the titration of 25.0 ml of 0.150 M nitrous acid (HNO2) with 0.150 M sodium hydroxide (NaOH), the pH at the equivalence point is calculated to be 8.11. At the equivalence point, the volumes of both solutions are equal, resulting in a total volume of 50.0 ml. To determine the new concentration, one must understand that both acid and base are present in equal proportions, leading to a neutralization reaction. The discussion emphasizes the importance of recognizing the definition of the equivalence point and calculating the new molarity of the resulting solution. This foundational understanding is crucial for accurately determining the pH after the titration.
rcrx
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A 25.0 ml sample of 0.150 M HNO2 is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of HNO2 is 4.50x10-4


What I can't get is the fact I don't know how much has already been titrated at the equivalence point. So it's difficult to know the total volume to find the new concentration.

The answer is supposed to be pH = 8.11, and I could guess that with the titration of a weak acid with a strong base that the pH will rise a bit. But I can't presume how to do this mathematically.
 
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rcrx said:
What I can't get is the fact I don't know how much has already been titrated at the equivalence point. So it's difficult to know the total volume to find the new concentration.
If you are having trouble with this, you should take a step back ... and start at the definition of the equivalence point. What is the definition?
 
Both concentrations are in equal proportion and neither consumes the other. So that means there is 25.0 + 25.0 for a combined total of 50ml?
 
Correct. Now, what's the next step?
 
Find the new Molarity of either substance?
 
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