Ka, Kb, calculating concentration of F-

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SUMMARY

The forum discussion focuses on calculating the equilibrium concentration of fluoride ions (F-) from a 0.158M solution of hydrofluoric acid (HF) using the provided base dissociation constant (Kb) of 6.66 x 10-4. Participants clarify that HF is a weak acid, which only partially dissociates in solution, and emphasize the importance of correctly applying the equilibrium expressions for Ka and Kb. The correct approach involves using the ICE (Initial, Change, Equilibrium) table method to derive the equilibrium concentrations accurately.

PREREQUISITES
  • Understanding of weak acids and their dissociation in water
  • Familiarity with the ICE table method for equilibrium calculations
  • Knowledge of the relationship between Ka, Kb, and Kw (Kw = Ka * Kb)
  • Ability to solve quadratic equations
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  • Learn how to derive Ka from Kb using the formula Ka = Kw / Kb
  • Study the concept of weak acid dissociation and its equilibrium expressions
  • Practice constructing and interpreting ICE tables for various acid-base reactions
  • Explore the implications of acid strength on dissociation and equilibrium concentrations
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Chemistry students, educators, and anyone involved in chemical equilibrium calculations, particularly in acid-base chemistry.

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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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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?
 

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