Ka, Kb, calculating concentration of F-

In summary, HF is a weak acid that only dissociates a little, and the equilibrium concentration of HF is 0.158M.
  • #1
ronpaulkid
8
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!
 
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  • #2
hey, I am new here. did i do something wrong?
 
  • #3
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.
 
  • #4
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?
 
  • #5
--------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?
 
  • #6
Initial is OK, but later you do some strange tricks.

Have you switched C & E lines?

Why do you think ALL HF was dissociated?
 

1. What is Ka and Kb?

Ka and Kb are equilibrium constants that represent the strength of an acid or base in a solution. Ka is used for acids and Kb is used for bases.

2. How do you calculate the concentration of F-?

The concentration of F- can be calculated using the formula [F-] = sqrt(Ka * c), where [F-] is the concentration of F-, Ka is the equilibrium constant for the acid, and c is the initial concentration of the acid.

3. What is the relationship between Ka and Kb?

Ka and Kb are related by the equation Ka * Kb = Kw, where Kw is the equilibrium constant for the autoionization of water. This means that as Ka increases, Kb decreases and vice versa.

4. How does temperature affect Ka and Kb?

Temperature has a direct effect on Ka and Kb. As temperature increases, the equilibrium constants also increase, indicating a stronger acid or base.

5. How do you determine the acid or base strength using Ka and Kb values?

The larger the value of Ka or Kb, the stronger the acid or base. A larger Ka indicates a stronger acid, while a larger Kb indicates a stronger base.

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