# Question 00d80

May 7, 2017

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 $\text{1 L}$ of solution. In your case, you have $1.3714$ moles of hydrochloric acid present in $\text{250 L}$ of solution, which means that $\text{1 L}$ 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 ${\text{0.0055 mol L}}^{- 1}$.

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

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

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 $\text{pH}$ of the solution is equal to

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

In your case, you will have

$\text{pH} = - \log \left(0.0055\right) = 2.3$

This value is rounded to one decimal place.