Question #00d80

1 Answer
May 7, 2017

Answer:

Here's what I got.

Explanation:

Start by calculating the number of moles of hydrogen chloride dissolved to make your solution.

#50 color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46 color(red)(cancel(color(black)("g")))) = "1.3714 moles HCl"#

Now, a solution's molarity tells you the number of moles of solute present for very #"1 L"# of solution. In your case, you have #1.3714# moles of hydrochloric acid present in #"250 L"# of solution, which means that #"1 L"# of solution will contain

#1 color(red)(cancel(color(black)("L solution"))) * "1.3714 moles HCl"/(250color(red)(cancel(color(black)("L solution")))) = "0.0055 moles HCl"#

You can thus say that the solution has a molarity of #"0.0055 mol L"^(-1)#.

As you know, hydrochloric acid si a strong acid, which implies that it ionizes completely in aqueous solution to produce hydrogen cations and chloride anions

#"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#

As you can see, every mole of hydrogen chloride that is dissolved in water produces #1# mole of hydrogen ions. This means that your solution will have

#["H"^(+)] = ["HCl"] = "0.0055 mol L"^(-1)#

I'll leave this value rounded to two sig figs, but keep in mind that you only have one significant figure for the mass of hydrogen chloride.

By definition, the #"pH"# of the solution is equal to

#"pH" = - log(["H"^(+)])#

In your case, you will have

#"pH" = - log(0.0055) = 2.3#

This value is rounded to one decimal place.