Question #85a11
1 Answer
Explanation:
The idea here is that you need to use the pH of the target solution to figure out the concentration of hydronium cations,
Once you know that, you can use the ionization of hydroiodic acid,
So, the pH of a solution is defined as the negative log of the concentration of hydronium cations
#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#
This means that the concentration of hydronium cations can be found by using the pH of the solution
#color(purple)(|bar(ul(color(white)(a/a)color(black)(["H"_3"O"^(+)] = 10^(-"pH"))color(white)(a/a)|)))#
In your case, the solution is said to have a pH equal to
#["H"_3"O"^(+)] = 10^(-3) = "0.0010 mol L"^(-1)#
Now, hydroionic acid is a strong acid, which means that it dissociates completely in aqueous solution to form hydronium cations and iodide anions,
#color(red)("H")"I"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(color(red)(+)) + "I"_((aq))^(-)#
This means that every mole of hydroionic acid placed in aqueous solution will form one mole of hydronium cations. Therefore, you can say that
#["HI"] = ["H"_3"O"^(+)]#
Since you know that the concentration of hydronium cations is given by the pH of the solution, you will have
#["HI"] = "0.0010 mol L"^(-1)#
As you know, a solution's molarity tells you how many moles of solute you get per liter of solution.
In this case, a molarity of
Since the problem tells you that you're dealing with one liter of this solution, you can say that it will contain
#"no. of moles of HI" = color(green)(|bar(ul(color(white)(a/a)0.0010 color(white)(a/a)|)))#
So, adding