Question

Question #00d80

Answer 1

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.