Ka, Kb, calculating concentration of F-

  • Thread starter Thread starter ronpaulkid
  • Start date Start date
  • Tags Tags
    Concentration
AI Thread Summary
To calculate the equilibrium concentration of F- from 0.158M HF, the correct approach involves using the relationship between Ka and Kb, where Ka can be derived from Kb. HF is identified as a weak acid, meaning it does not fully dissociate, contrary to the initial assumption of it being a strong acid. The ICE table method is recommended for determining concentrations at equilibrium, and the initial concentration should not be confused with equilibrium values. The calculations must accurately reflect the dissociation of HF, and there may be confusion regarding the values of Ka and Kb. Properly setting up the ICE table and equations is crucial for arriving at the correct concentration of F-.
ronpaulkid
Messages
8
Reaction score
0

Homework Statement



Calculating concentration of F- for .158M of HF?
Calculate the equilibrium concentration of F- for .158M of HF in the reaction below.?
HF+H20=H30+F-

The Kb is 6.66*10-4

Homework Equations



Kw=Ka*Kb
Ka=[H+][A-]/[HA]


The Attempt at a Solution



I figured that you would find the Ka by taking 1*10^-14 over the Kb.
Since Ka=[H+][A-]/[HA], then Ka=x^2/.158M

However, the answer is not 1.54E-6.

I also made an ICE chart and solving for x using the quadratic formula. I must be missing something. Thanks!
 
Physics news on Phys.org
hey, I am new here. did i do something wrong?
 
Your equation is ambigious, there is no equal sign in a chemical reaction.

If I remember, HF is a strong acid, it only proceeds in one direction.
 
ronpaulkid said:
hey, I am new here. did i do something wrong?

Yes - this is not physics, this is chemistry, and chemistry - when it comes to homework - classifies as other sciences...

flyingpig said:
Your equation is ambigious, there is no equal sign in a chemical reaction.

This is a perfect shortcut, especially for an equilibrium reaction.

If I remember, HF is a strong acid, it only proceeds in one direction.

HF is a weak acid. Besides, even strong acids are never 100% dissociated (although in most cases that's a perfect approximation).

ronpaulkid said:
Since Ka=[H+][A-]/[HA], then Ka=x^2/.158M

0.158M is a formal concentration of acid, not equilibrium concentration of HF.

I also made an ICE chart and solving for x using the quadratic formula. I must be missing something. Thanks!

Show your ICE table, it should yield a correct result.

Edit: actually it seems like acid is dissociated only slightly, even first approach should give answer reasonably close to reality. Are you sure 6.66*10-4 is Kb, and not Ka?
 
--------HF+H20=F-+H30
I-----.158---0-----0-----0

C-----(-.158)--0---.158---0

E------x-------0------(-x)----0

Ka=x^2/[x-.158]

Kb=[x-.158]/x^2

Is this close to how it looks?
 
Initial is OK, but later you do some strange tricks.

Have you switched C & E lines?

Why do you think ALL HF was dissociated?
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top