# What is the pH of a solution in which 1\times10^(-7) moles of the strong acid, HCl is added to one liter of water?

## I am going to put what I think is the answer, please correct my solution if it's wrong finding molarity: $\frac{1 \setminus \cdot {10}^{- 7} m o l}{1 L} = 1 \setminus \cdot {10}^{- 7} M$ R) $H C l \setminus r i g h t \le f t h a r p \infty n s$..... ${H}^{+} + C l -$ I) $1 \setminus \cdot {10}^{- 7} M$ ...... 0 ....... 0 C) ...... $- x$ ......... ... $x$ ...... $x$ E) $1 \setminus \cdot {10}^{- 7} - x$ .... $x$ .... $x$ I don't know where to go from here as I don't know what ${K}_{a}$ is. I would apply the Henderson-Hasselbalch equation, but that's the next chapter down the line... and I don't have enough information for that anyway.

Mar 16, 2017

$\text{pH} = 6.8$

#### Explanation:

Your starting point here will be the auto-ionization of water, which looks like this

$2 {\text{H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

At room temperature, water has an ionization constant, ${K}_{W}$, equal to

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

Now, you know that hydrochloric acid is a strong acid, which implies that it ionizes completely in aqueous solution to produce hydronium cations

${\text{HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

Notice that the concentration of hydronium cations is equal to the initial concentration of the acid, which in your case is

(1 * 10^(-7)"moles HCl")/"1 L solution" = 1 * 10^(-7)color(white)(.)"M"

The trick now is to realize that after you add the moles of acid, the auto-ionization equilibrium still takes place!

In other words, after you add $1 \cdot {10}^{- 7} \text{M}$ of hydronium cations, the auto-ionization equilibrium will produce $x$ $\text{M}$ of hydronium cations and $x$ $\text{M}$ of hydroxide anions.

So you can say that after adding the acid, you have

${K}_{W} = {\overbrace{\left(x + 1 \cdot {10}^{- 7}\right)}}^{\textcolor{b l u e}{{\text{the total concentration of H"_3"O"^(+))) * overbrace(" "x" ")^(color(purple)("the total concentration of OH}}^{-}}} = {10}^{- 14}$

This means that you have

${x}^{2} + 1 \cdot {10}^{- 7} x - {10}^{- 14} = 0$

This quadratic equation will produce two values for $x$, a positive value and a negative value. Since $x$ represents concentration, the negative value will hold no physical significance.

You will thus have

$x = 6.18 \cdot {10}^{- 8}$

This means that the total concentration of hydronium cations in the final solution will be

["H"_3"O"^(+)] = 6.18 * 10^(-8)"M" + 1 * 10^(-7)"M" = 1.618 * 10^(-7)"M"

As you know, the pH of the solution is given by

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

In your case, you will have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = - \log \left(1.618 \cdot {10}^{- 7}\right) = 6.8}}}$

The value must be rounded to one decimal place, the number of sig figs you have for the number of moles of hydrogen chloride dissolved to make the solution of hydrochloric acid.

Now, does the result make sense?

Hydrochloric acid is a strong acid, so even in this very, very small concentration, the pH of the solution must come out to be $< 7$.