# Why is water ignored in the ionic equilibrium of HF?

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 $$x$$?

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 $$\ce{H2O}$$?
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?

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:

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