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

#"pH" = 1.5#


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.