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

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SUMMARY

The discussion focuses on calculating the formate ion concentration and pH of a mixed solution containing 0.05 M formic acid (HCOOH) and 0.10 M nitric acid (HNO3). Participants clarify that nitric acid, being a strong acid, completely dissociates in solution, contributing significantly to the hydronium ion concentration. The correct approach involves recognizing that the pH is primarily determined by the strong acid, leading to a pH of approximately 1, while the weak acid's dissociation can be neglected due to the dominance of HNO3. The calculations presented demonstrate the importance of understanding acid strength and dissociation in mixed solutions.

PREREQUISITES
  • Understanding of acid-base chemistry, specifically strong vs. weak acids.
  • Familiarity with the dissociation equations for acids.
  • Knowledge of the acid dissociation constant (Ka) and its application.
  • Ability to perform logarithmic calculations for pH determination.
NEXT STEPS
  • Study the dissociation constants of various weak acids, including formic acid (HCOOH).
  • Learn about the common ion effect and its impact on weak acid dissociation.
  • Explore the concept of strong acid dissociation and its implications in mixed solutions.
  • Review pH calculation methods for solutions containing both strong and weak acids.
USEFUL FOR

Chemistry students, educators, and professionals involved in acid-base chemistry, particularly those working with mixed solutions and pH calculations.

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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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