Question

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?

Answer 1

`"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 `"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 `"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 "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.

`"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)`

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 `"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

`"pH" = - log(0.02857) = color(darkgreen)(ul(color(black)(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.