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While trying to understand the solution of a problem given in my text book, I realized I'm having some difficulty with the solution. The problem is as folows:

The ionization constant of HF is $3.2 \times 10^{-4}$. Calculate the degree of dissociation of HF in its 0.02 solution. Calculate the concentration of all species present $\ce{H3O+ , F+}$ and $\ce{HF}$ in the solution and its PF.

In the solution of this problem, the equation is given as

$$\ce{HF + H2O <-> H3O+ + F-}$$

The concentration of at the time of equilibrium are given as:

$$[\ce{HF}] = 0.02 - 0.02x, \: \ce{[H3O+]} = 0.02x, \:\ce{[F- ]}= 0.02x$$

I have the following doubts:

  1. Why we are not adding the contribution of water to the $\ce{H3O+}$ ions?
  2. Why we are assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$.?
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  • $\begingroup$ 1. you need units. There's a difference between (e.g.) 0.02 mol/l, 0.02 mmol/l, 0.02 = 2 %, ... $\endgroup$ – cbeleites supports Monica May 24 '12 at 17:42
  • $\begingroup$ 2. Your question sounds homeworky to me. Even if it is not really homework that was assigned to you, I'd suggest you tag it as homework as this will tell others what kind of question to expect, and possibly also change the answers to something that helps you educating yourself. $\endgroup$ – cbeleites supports Monica May 24 '12 at 17:44
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(Assuming this is homework or self-education which should be treated homeworky).

1) why we are not adding the contribution of water to the H3O+ ions

  1. Write down the dissociation constant equation with and without $\ce{H2O}$. What changes?
  2. Can you find a formulation of the dissociation constant where it seems natural not to include $\ce{H2O}$?
  3. Read up "activity".

2) why we are assuming that the value 0.02x is dissociated from HF and not just x

Write down & solve the equation with 0.02 x and with x only. What is the difference?

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1) why we are not adding the contribution of water to the H3O+ ions

$\ce{H3O+}$ actually is $\ce{H+}$ and $\ce{H2O}$ so you can write the equation as just dissociation of $\ce{HF}$ ($\ce{H2O}$ cancels out on both sides)
$$\ce{HF <=> H+ + F-}$$ Thus we do not take water into consideration.

2) why we are assuming that the value 0.02x is dissociated from HF and not just x

Because whenever we write an equation
$$\ce A \ce{->} \ce B + \ce C $$ We write initial concentration of A as $c$ while B & C are zero. Final concentrations are written as

  • for A: (Initial concentration) - (concentration dissociated)
  • for B & C: (concentration dissociated)

Thus re-writing the equation

               HF <-------> H+ + F-  
Initial Moles: 0.2          0    0  
Final Moles:   0.2-0.2x    .2x  .2x   
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  • 2
    $\begingroup$ I've heavily edited your answer so it's more readable. A few things that were incorrect: 1. a mix of quantities (incorrectly named “moles”, which is the unit) and concentrations, 2. abbreviations throughout, 3. chemical species and equations formatting (proper subscripts, arrow, etc.). Please try to keep an eye out for that in future answers. $\endgroup$ – F'x May 24 '12 at 19:01
  • $\begingroup$ @F'x Sir im sorry for that. I was feeling a bit lazy but was going to come up later and do the editing.. if in future you have any problem with any of my posts please post a comment. Dont take the trouble to edit it. Sorry again. $\endgroup$ – Ashu May 26 '12 at 17:38
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    $\begingroup$ @Ashu, this is a community-driven site, and community editing is an integral part of that. Don't feel bad if others take time out to edit your posts--if a post can be improved, we will try to get it done. It's not "trouble" for us :) . Though if you know that your post is not that good but can be easily improved, improve it before submitting. Note that the site saves drafts for you, so you can come back if you feel too lazy to do it now :) $\endgroup$ – ManishEarth May 26 '12 at 19:11

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