Question #e049b
1 Answer
Explanation:
The trick here is to realize that hydrochloric acid,
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
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
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)|)))