# A chemist dissolves 0.01 mol of an acid (HCl) in enough water to make a 350 mL solution of acid. What is the pH of the acid solution?

##### 1 Answer
Mar 7, 2018

$\text{pH} = 1.5$

#### Explanation:

For starters, you need to calculate the molar concentration of the acid by figuring out how many moles of hydrochloric acid would be present in $\text{1 L" = 10^3 quad "mL}$ of this solution.

To do that, use the fact that you dissolved $0.01$ moles of hydrochloric acid to make $\text{350 mL}$ of the solution.

10^3 color(red)(cancel(color(black)("mL solution"))) * "0.01 moles HCl"/(350color(red)(cancel(color(black)("mL solution")))) = "0.02857 moles HCl"

This means that the molar concentration of the solution is equal to

["HCl"] = "0.02857 mol L"^(-1)

This implies that ${10}^{3} \quad \text{mL" = "1 L}$ of this solution contains $0.02857$ moles of hydrochloric acid.

Now, hydrochloric acid is a strong acid, which implies that it dissociates completely in aqueous solution to produce hydronium cations and chloride anions.

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

This means that all the moles of hydrochloric acid that you add to the solution will dissociate to produce hydronium cations.

Therefore, you can say that your solution has

["H"_3"O"^(+)] = ["HCl"] = "0.02857 mol L"^(-1)

As you know, the $\text{pH}$ of the solution is calculated by using the equation

color(blue)(ul(color(black)("pH" = - log(["H"_ 3"O"^(+)]))))

Plug in your value to find

$\text{pH} = - \log \left(0.02857\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.5}}}$

The answer is rounded to one decimal place, the number of sig figs you have for the number of moles of hydrochloric acid present in the solution.