# Calculate the "pH" of 10^-8 "M" "HCl"?

Jan 31, 2016

$\text{pH} = 6.98$

#### Explanation:

This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.

As you know, pure water undergoes self-ionization to form hydronium ions, ${\text{H"_3"O}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$.

$\textcolor{p u r p \le}{2 {\text{H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}} \to$ very important!

At room temperature, the value of water's ionization constant, ${K}_{W}$, is equal to ${10}^{- 14}$. This means that you have

${K}_{W} = \left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right] = {10}^{- 14}$

Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have

["H"_3"O"^(+)] = sqrt(10^(-14)) = 10^(-7)"M"

The pH of pure water will thus be

color(blue)("pH" = - log(["H"_3"O"^(+)]))

$\text{pH} = - \log \left({10}^{- 7}\right) = 7$

Now, let's assume that you're working with a $\text{1.0-L}$ solution of pure water and you add some ${10}^{- 8} \text{M}$ hydrochloric acid solution.

To keep calculations simple, let's assume that the volume remains unchanged upon adding this hydrochloric acid solution.

Hydrocloric acid is a strong acid, which means that it dissociates completely to form hydronium ions and chloride anions.

${\text{HCl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

Since you have a $1 : 1$ mole ratio between the acid and the hydronium ions, you will have

["H"_3"O"^(+)] = ["HCl"] = 10^(-8)"M"

The absolute most important thing to realize now is that adding acid to the pure water does not disrupt water's self-ionization reaction!

Water will still self-ionize to produce equal concentrations of hydronium and hydroxide ions. However, you should expect the excess of hydronium ions to actually decrease the number of moles of ions produced by the self-ionization reaction.

Let's say that After the acid has been added, water's self-ionization produces $x$ moles of hydronium ions and $x$ moles of hydroxide ions. You can say that

$\left(x + {10}^{- 8}\right) \cdot x = {10}^{- 14}$

Here $\left(x + {10}^{- 8}\right)$ represents the equilibrium concentration of hydronium ions and $x$ represents the equilibrium concentration of hydroxide ions.

Remember, we're using a $\text{1.0-L}$ sample, so moles and concentration are interchangeable.

Rearrange this equation to solve for $x$

${x}^{2} + {10}^{- 8} x - {10}^{- 14} = 0$

This quadratic equation will produce two values for $x$. Since $x$ represents concentration, you must pick the positive one

$x = 9.51 \cdot {10}^{- 8}$

Therefore, the equilibrium concentration of hydronium ions will be

["H"_3"O"^(+)] = 10^(-8) + 9.51 * 10^(-8) = 1.051 * 10^(-7)"M"

The pH of the solution will thus be

$\text{pH} = - \log \left(1.051 \cdot {10}^{- 7}\right) = \textcolor{g r e e n}{6.98}$

And remember, the pH of an acidic solution can never, ever, be higher than $7$!