# Question 85a11

Apr 16, 2016

$\text{0.0010 moles HI}$

#### Explanation:

The idea here is that you need to use the pH of the target solution to figure out the concentration of hydronium cations, ${\text{H"_3"O}}^{+}$, it must contain.

Once you know that, you can use the ionization of hydroiodic acid, $\text{HI}$, and the volume of the solution to figure out the number of moles needed.

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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, the solution is said to have a pH equal to $3$, which means that the concentration of hydronium ions will be

["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, ${\text{I}}^{-}$, in $1 : 1$ mole ratios

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

$\left[{\text{HI"] = ["H"_3"O}}^{+}\right]$

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 ${\text{0.0010 mol L}}^{- 1}$ tells you that one liter of this solution will contain $0.0010$ moles of hydroionic acid, your solute.

Since the problem tells you that you're dealing with one liter of this solution, you can say that it will contain

$\text{no. of moles of HI} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 0.0010 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, adding $0.0010$ moles of $\text{HI}$ in enough water to make one liter of solution will get a solution that has a pH of $3$.