Question #e049b

The trick here is to realize that hydrochloric acid, #"HCl"#, is a strong acid, which implies that it ionizes completely in aqueous solution to form hydronium cations, #"H"_3"O"^(+)#, sometimes referred to as hydrogen ions, #"H"^(+)#, and chloride anions, #"Cl"^(-)#.

Since every molecule of hydrogen chloride contains one atom of hydrogen and one atom of chlorine, it follows that one mole of hydrochloric acid will ionize to form

• one mole of hydronium cations, #1 xx "H"_3"O"^(+)#
• one mole of chloride anions, #1 xx "Cl"^(-)#

You will thus have

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

Use the molarity and volume of the initial solution to determine how many moles of hydrochloric acid it contains.

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

You will have

#n_(HCl) = "0.025 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(15 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#

#n_(HCl) = 3.75 * 10^(-4)"moles HCl"#

Since you know that hydrochloric acid ionizes in a #1:1# mole ratio to form hydronium cations, you can say that the initial solution contains

#n_(H_3O^(+)) = color(green)(|bar(ul(color(white)(a/a)3.8 * 10^(-4)"moles H"_3"O"^(+)color(white)(a/a)|)))#

The answer is rounded to two sig figs.

Now, you add #"25 mL"# to this initial solution. The total volume will now be

#V_"total" = "15 mL" + "25 mL" = "40. mL"#

Since this solution contains the same number of moles of hydronium cations as the initial solution, it follows that the concentration of the hydronium cations, which is equivalent to that of the hydrochloric acid, will be

#["H"_3"O"^(+)] = (3.75 * 10^(-4)"moles")/(40. * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"0.0094 M"color(white)(a/a)|)))#

Once again, the answer is rounded to two sig figs.

To find the pH of the solution, use the equation

#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#

Plug in the concentration of hydronium cations to get

#"pH" = - log(0.0094) = color(green)(|bar(ul(color(white)(a/a)2.03color(white)(a/a)|)))#