What is the ph of a 100 ml solution containing 0.0040 g of HCl?

Dec 19, 2016

$\text{pH} = 3.0$

Explanation:

The first thing to do here is to calculate the number of moles of hydrochloric acid present in that sample. To do that, use the compound's molar mass

0.0040 color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46 color(red)(cancel(color(black)("g")))) = "0.0001097 moles HCl"

Now, hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to produce hydronium cations, ${\text{H"_3"O}}^{+}$

${\text{HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

This basically means that every mole of hydrochloric acid dissolved in water will produce $1$ mole of hydronium cations.

$\text{no. of moles of H"_ 3"O"^(+) = "0.0001097 moles}$

The molarity of the hydronium cations must be calculated for $\text{1 L}$ of solution, so do

1 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * ("0.0001097 moles H"_3"O"^(+))/(100color(red)(cancel(color(black)("mL solution"))))

$= {\text{0.001097 moles H"_3"O}}^{+}$

Since that is how many moles of hydronium cations you have in $\text{1 L}$ of solution, you can say that the concentration of the hydronium cations will be

["H"_3"O"^(+)] = "0.001097 M"

The pH of the solution is calculated by taking the negative log of the concentration of hydronium cations

color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))

Plug in your value to find

$\text{pH} = - \log \left(0.001097\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{3.0}}}$

The answer is rounded to one decimal place, the number of sig figs you have for the volume of the solution.