# What is the "pH" for "0.11 M" "HF" dissociating in water at 25^@ "C"? K_a = 7.2 xx 10^(-4)

## Is there enough information to do this?

##### 1 Answer
Jun 8, 2017

I got $2.07$.

Well, we have enough information, so yes.

The main things you'll need to know how to do:

1. Write a weak acid dissociation reaction of $\text{HF}$ in water.
2. Write an ICE table underneath it and fill it out.
3. Write the mass action expression for ${K}_{a}$, the acid dissociation constant.
4. Solve for $x$, which tends to require the quadratic formula. In this case, $x$ represents $\left[{\text{H"_3"O}}^{+}\right]$.
5. Use $\left[{\text{H"_3"O}}^{+}\right]$ to get $\text{pH}$.

1)

${\text{HF"(aq) + "H"_2"O"(l) rightleftharpoons "F"^(-)(aq) + "H"_3"O}}^{+} \left(a q\right)$

Since $\text{HF}$ is an acid, it donates an ${\text{H}}^{+}$ to water, and becomes its conjugate base, ${\text{F}}^{-}$.

2)

ICE Tables are a way to organize the $\boldsymbol{\text{I""nitial}}$, $\boldsymbol{\text{C""hange}}$, and $\boldsymbol{\text{E""quilibrium}}$ concentrations for the reactants and products.

When you are given a concentration of a weak acid or base with no other information, it tends to be the initial concentration.

${\text{HF"(aq) " "+" " "H"_2"O"(l) " "rightleftharpoons" " "F"^(-)(aq) + "H"_3"O}}^{+} \left(a q\right)$

$\text{I"" ""0.11 M"" "" "" "" "-" "" "" "" "" ""0 M"" "" "" ""0 M}$

$\text{C"" "-x" "" "" "" "" "-" "" "" "" "+x" "" "" } + x$

$\text{E"" "(0.11 - x)"M"color(white)(.....)-" "" "" "" "x" M"" "" "" "x" M}$

If any stoichiometric coefficients were greater than $1$, you would see that coefficient in front of the $x$ for that component in the reaction.

• Reactants have negative changes in concentration towards equilibrium (because they get consumed!).
• Products have positive changes in concentration (they get... produced!).

So, if you saw $2 \text{HF}$ for a reactant, you would put $- 2 x$ for the change. If you saw $3 {\text{OH}}^{-}$ for a product, you would put $+ 3 x$ for the change.

3)

The mass action expression is just the equilibrium constant as a function of the equilibrium concentrations, with products over reactants, raised to their respective stoichiometric coefficients:

${K}_{a} = \left(\left[\text{H"_3"O"^(+)]["F"^(-)])/(["HF}\right]\right)$

$= \frac{x \cdot x}{0.11 - x} = {x}^{2} / \left(0.11 - x\right)$

4) It's up to you how you want to use ${K}_{a} = 7.2 \times {10}^{- 4}$ to solve this:

• Solve for the quadratic equation form, $a {x}^{2} + b x + c = 0$ and use the quadratic formula to solve this in full.
• Use the small $\boldsymbol{x}$ approximation to cross out the $x$ for the weak acid reactant (or base, if it's a weak base equilibrium). This works when $K$ is small enough, and if it's ${10}^{- 5}$ or less, chances are that it'll work well.
• Use the "improved" small $\boldsymbol{x}$ approximation. This is what I'll do here, and I go into it in more detail here:

${K}_{a} \approx {x}^{2} / 0.11$

=> x^"*" ~~ sqrt(K_a(0.11 - x)) -= ["H"_3"O"^(+)]

Using the third way, what we'll do is set the first $x$ to be $0$, and acquire our first ${x}^{\text{*}}$. Our first "guess" is:

${x}_{1} = \sqrt{7.2 \times {10}^{- 4} \left(0.11\right)} = \text{0.008899 M}$

Now what we do is recycle ${x}_{1}$ as $x$ and get another ${x}^{\text{*}} = {x}_{2}$:

${x}_{2} = \sqrt{7.2 \times {10}^{- 4} \left(0.11 - 0.008899\right)}$

$=$ $\text{0.008532 M}$

And if we repeat this several more times:

${x}_{3} = \sqrt{7.2 \times {10}^{- 4} \left(0.11 - 0.008532\right)}$

$=$ $\text{0.008547 M}$

${x}_{4} = \sqrt{7.2 \times {10}^{- 4} \left(0.11 - 0.008547\right)}$

$=$ $\text{0.008547 M}$

So, our answer has converged upon ["H"_3"O"^(+)] = "0.008547 M". This works when $\frac{{K}_{a}}{\left[\text{HA}\right]} < 1$, and indeed:

$\frac{7.2 \times {10}^{- 4}}{0.11} \text{<<} 1$.

5)

So, the $\text{pH}$ is:

$\textcolor{b l u e}{{\text{pH") = -log["H"_3"O}}^{+}}$

$= - \log \left(\text{0.008547 M"//"1 M}\right)$

$= \textcolor{b l u e}{2.07}$