How Do You Calculate Formate Ion Concentration and pH in Mixed Solutions?

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To calculate the formate ion concentration and pH of a solution containing 0.05 M formic acid and 0.10 M nitric acid, it's essential to recognize that nitric acid is a strong acid that fully dissociates, contributing significantly to hydronium ion concentration. The dissociation of formic acid, a weak acid, is influenced by the presence of the strong acid, which may suppress its dissociation. The correct approach involves using the strong acid's concentration to determine the pH, which is primarily dictated by the nitric acid. The calculations indicate that the pH is closer to 1 due to the dominant contribution of nitric acid. Understanding the interplay between the strong and weak acids is crucial for accurate pH determination.
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Homework Statement


calculate the formate ion concentration and pH of a solution that is 0.05m in formic acid (HCOOH;Ka=1.8x10^-4) and 0.10M in HNO3.


hey guys just wondering how the equation goes??

like this?? HCOOH + HNO3>>?
 
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Would knowing that nearly all of the hydronium ion concentration is from the nitric acid help you?
 
The equation you need is the equation for Ka. Try that one first...
 
when i tried doing the Ka i get x/(0.05*0.1)=1.8e-4 therefore x=3.6e-4 however when i -log3.6e-4 i get a ph of 3.44, however the answer is a pH of 1?? thanks for the replys
 
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Nitric Acid is a strong acid; this means that its ions completely dissociate in water solution. You essentially have a solution which has 0.10 M in hydronium ion. What would be the pH of just that? How would 0.10 M hydronium affect the dissociation of the formic acid?
 
does that mean just make the dissocation equation just with HNO3. well this is what i did. HNO3>NO3 + H therefore equation equals. [H][NO3]/[HNO3]=1.8e-4 therefore solving this x^2/0.1=1.8e-4 therefore i get a ph of 2.37? thansk for replying
 
geffman1 said:
does that mean just make the dissocation equation just with HNO3. well this is what i did. HNO3>NO3 + H therefore equation equals. [H][NO3]/[HNO3]=1.8e-4 therefore solving this x^2/0.1=1.8e-4 therefore i get a ph of 2.37? thansk for replying

You are misapplying your information. Nitric acid is a strong acid. Formic acid is the weak acid. The concentration of nitric acid is high enough that it may push the dissociation of formic acid to the left in regard to HCHO <------> H+ + CHO-
 
i am confused haha, could u show me ur working for the dissocaition working please, if it is a weak acid isn't that's y the give the ka value for it. so does this just mean i used the dissociation expression for HCOOH and just used -log for HNO3? thansk
 
wow i think it just clicked, since HCOOH is weak, can we ignored it and just use the concentration of NO3 because its strong. because that works out. if that is right u just saved me so much time. thanksss heaps
 
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Fine! This indicates some good progress.
 

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