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

The ionization constant of $\ce{HF}$ is $3.2 \times 10^{-4}$. Calculate the degree of dissociation of $\ce{HF}$ in its $\pu{0.02 M}$ solution. Calculate the concentration of all species present $\ce{H3O+}$, $\ce{F-}$ 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 at the time of equilibrium are given as:

\begin{align} \ce{[HF]} &= 0.02 - 0.02x, & \ce{[H3O+]} &= 0.02x, & \ce{[F- ]} &= 0.02x \end{align}

I have the following questions:

  1. Why are we not adding the contribution of water to the $\ce{H3O+}$ ions?
  2. Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?
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  1. Why are we not adding the contribution of water to the $\ce{H3O+}$ 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 on "activity".
  1. Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?

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

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  1. Why are we not adding the contribution of water to the $\ce{H3O+}$ 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.

  1. Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?

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

  • for $\ce{A}$: (Initial concentration) - (concentration dissociated)
  • for $\ce{B}$ and $\ce{C}$: (concentration dissociated)

Thus re-writing the equations:

\begin{array}{lccccc} & \ce{HF} & \ce{<=>} & \ce{H+} & + & \ce{F-} \\ \text{Initial Amount}:& 0.2 && 0 && 0 \\ \text{Final Amount}: & 0.2-0.2x && 0.2x && 0.2x \\ \end{array}

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