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

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, #"H"_3"O"^(+)#

#"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#

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

#"no. of moles of H"_ 3"O"^(+) = "0.0001097 moles"#

The molarity of the hydronium cations must be calculated for #"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"))))#

#= "0.001097 moles H"_3"O"^(+)#

Since that is how many moles of hydronium cations you have in #"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

#"pH" = - log(0.001097) = color(darkgreen)(ul(color(black)(3.0)))#

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