# A certain sample of "HClO"_2 has a concentration of "0.125 M". Calculate the pH at equilibrium. Why can we not treat "HClO"_2 as a strong acid, even though K_a > 1?

Mar 4, 2016

$\text{pH}$ $= 1.496 \approx 1.50$

#### Explanation:

$H C l {O}_{2}$ is a weak acid. You can write the equation like this:

$H C l {O}_{2} + {H}_{2} O r i g h t \le f t h a r p \infty n s C l {O}_{2}^{-}$$+$${H}_{3} {O}^{+}$
(Because $H C l {O}_{2}$ is an acid, it donates a hydrogen)

Because the $H C l {O}_{2}$ is a weak acid, it does not dissociate/ionize completely.
To find the pH, we will have to do problem solving similar to equilibrium problems.
We will use the ICE table first of all:

-------- $H C l {O}_{2}$ -- ${H}_{2} O$ $r i g h t \le f t h a r p \infty n s$ $C l {O}_{2}$ -- ${H}_{3} {O}^{+}$

Initial: 0.125M -- N/A --------- 0M --- 0M
Change: x ------- N/A ---------- x ------ x
Equilib.: 0.125-x -- N/A ------- x ------ x

The next step is to find out what x is, and we do this by using the "equilibrium equation" but with Ka instead of just K.

The equation for this is:

$K a = \left(\left[\text{products"])/(["reactants}\right]\right)$

Which is:

$K a = \frac{\left[{H}_{3} O\right] \left[C l {O}_{2}\right]}{H C l {O}_{2}}$

To solve this, you will have to look up the value of Ka for $H C l {O}_{2}$, which is $1.1 \cdot {10}^{-} 2$.

Now, plug in the equilibrium values from the ICE table:

$1.1 \cdot {10}^{-} 2$= $\frac{\left[x\right] \left[x\right]}{\left[0.125 - x\right]}$

Okay, so now what's left is algebra, to find out the value of x.
After all the algebra, we are left with the quadratic equation of:

${x}^{2} - 1.375 \cdot {10}^{-} 3 + 1.1 \cdot {10}^{-} 2 x$.

**Note: in some cases, typically where Ka is VERY small, you can assume that x (in the 0.125M-x part) is so small in relation, that you can ignore it completely and not go through the trouble of using the quadratic formula.

To solve for x from this, we'll use the quadratic formula, so:

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\frac{- 1.1 \cdot {10}^{-} 2 \pm \sqrt{{\left(1.1 \cdot {10}^{-} 2\right)}^{2} - 4 \left(1\right) \left(- 1.375 \cdot {10}^{-} 3\right)}}{2 \cdot 1}$

$= \frac{- 1.1 \cdot {10}^{-} 2 \pm 0.075}{2}$

You want what will give you a positive number, so:

$\frac{- 1.1 \cdot {10}^{-} 2 + 0.075}{2}$

$x = 0.032$

$\left[{H}_{3} {O}^{+}\right]$ is equal to x, so pH is equal to the -log of the hydronium concentration (written as ${H}^{+} \mathmr{and} {H}_{3} {O}^{+}$)

$p H = - \log \left[{H}_{3} {O}^{+}\right]$

$p H = - \log \left[0.032\right]$

$\textcolor{b l u e}{p H = 1.496}$

Mar 5, 2016

${\text{HClO}}_{2}$ has a pKa is about $1.96$, so one might think that it is a fairly strong acid in water. Let's start with the assumption that it is "strong" and compare it to if it is "weak". Then we can see whether it matters. I think it will matter.

In general, ${\text{H"_3"O}}^{+}$ is given ${K}_{a} = 1$, and a good rule of thumb is that the ${K}_{a}$ should be at least $10 - 100$ times larger to have a strong acid.

ASSUMPTION: "STRONG" ACID

If it is "strong", then "pH" = -log["H"^(+)], where we say that $\left[{\text{H"^(+)] = ["HClO}}_{2}\right]$, so

$\textcolor{red}{\text{pH") = -log("0.125 M}}$

$= 0.903090 \approx \textcolor{red}{0.903}$

ASSUMPTION: "WEAK" ACID

Alright, now let's say it was weak.

${K}_{a} = {10}^{- \text{pKa}} = {10}^{- 1.96} \approx 0.01096$

Now we'll write out the equilibrium reaction and ICE table for it:

${\text{HClO"_2(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "ClO}}_{2}^{-} \left(a q\right)$

$\text{I"" "0.125" "" "" "-" "" "" "" "0" "" "" "" } 0$
$\text{C"" "-x" "" "" "color(white)(.)-" "" "" "+x" "" "" } + x$
$\text{E"" "0.125-x" "-" "" "" "" "x" "" "" "" } x$

The ${\text{K}}_{a}$ can be utilized:

${K}_{a} = \left(\left[{\text{H"_3"O"^(+)]["ClO"_2^(-)])/(["HClO}}_{2}\right]\right)$

When you examine the ICE table, you should see that the equilibrium concentrations of each reactant are:

$\left[{\text{H"_3"O"^(+)] = ["ClO}}_{2}^{-}\right] = x$
$\left[{\text{HClO}}_{2}\right] = 0.125 - x$

It's really a generalized way of saying "It started at this concentration and dissociated into the same molar ratio of proton to conjugate base, and the acid's equilibrium concentration is that much less."

Hence:

${K}_{a} = \frac{{x}^{2}}{0.125 - x}$

This cannot be used with the small x approximation because ${K}_{a}$ is too big. The resulting quadratic is:

$0.01096 \cdot 0.125 - 0.01096 x - {x}^{2} = 0$

After we use the quadratic formula, we would get the positive $x$ to be:

$x = \text{0.031937 M}$

That gives an ${\text{H}}^{+}$ concentration of $\text{0.031937 M}$, which means:

$\textcolor{b l u e}{\text{pH") = -log("0.031937 M}}$

$= 1.496 \approx \textcolor{b l u e}{1.50}$

Clearly we get drastically different answers in each approach, so it definitely matters, and we couldn't have assumed that chlorous acid was a strong acid.