Question

Question #9190f

Answer 1

`"pH" = 2`

Explanation:

You're dealing with two strong acids, hydrochloric acid, `"HCl"`, and nitric acid, `"HNO"_3`, and a strong base, sodium hydroxide, `"NaOH"`, so right from the start you know that both acids and the base will be completely ionized in the resulting solution.

Both acids dissociate in a `1:1` mole ratio to form hydronium cations, `"H"_3"O"^(+)`

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

`color(red)("H")"NO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(color(red)(+)) + "NO"_(3(aq))^(-)`

The base dissociates in a `1:1` mole ratio to form hydroxide anions, `"OH"^(-)`

`"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) +"OH"_ ((aq))^(-)`

Your strategy here will be to use the molarities and volumes of the three solutions to find how many moles of hydronium cations and how many moles of hydroxide anions are mixed

`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)|)))`

Keep in mind that you have

`"1 dm"^3 = 10^3"cc" = 10^3"cm"^3`

The number of moles of hydronium cations coming from the hydrochloric acid solution will be

`n_("H"_3"O"^(+)"/ HCl") = 1/3color(white)(a)"mol" color(red)(cancel(color(black)("dm"^(-3)))) * 30 * 10^(-3)color(red)(cancel(color(black)("dm"^3))) = "0.010 mol H"_3"O"^(+)`

The number of moles of hydronium cations coming from the nitric acid solution will be

`n_("H"_3"O"^(+)"/ HNO"_3) = 1/2color(white)(a)"mol" color(red)(cancel(color(black)("dm"^(-3)))) * 20 * 10^(-3)color(red)(cancel(color(black)("dm"^3))) = "0.010 mol H"_3"O"^(+)`

The total number of moles of hydronium cations present in solution will be

`n_("H"_3"O"^(+)" total") = "0.010 moles" + "0.010 moles" = "0.020 moles H"_3"O"^(+)`

The number of moles of hydroxide anions coming from the sodium hydroxide solution will be

`n_("OH"^(-)"/ NaOH") = 1/4color(white)(a)"mol" color(red)(cancel(color(black)("dm"^(-3)))) * 40 * 10^(-3)color(red)(cancel(color(black)("dm"^3))) = "0.010 mol OH"^(-)`

The hydronium cations and the hydroxide anions will neutralize each other to form water

`"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))`

Notice that the two ions react in a `1:1` mole ratio; in your case, you have fewer moles of hydroxide anions than of hydronium cations, so you know that the hydroxide anions will act as a limiting reagent.

That it, they will be completely consumed by the reaction. After the neutralization reaction is finished, the resulting solution will contain

`n_("OH"^(-)) = "0 moles OH"^(-) ->` completely consumed

`n_("H"_3"O"^(+)) = "0.020 moles" - "0.010 moles" = "0.010 moles H"_3"O"^(+)`

You know that the total volume of the solution was made up to `"1 dm"^3`, which means that the concentration of hydronium cations in the resulting solution will be

`["H"_3"O"^(+)] = "0.010 moles"/"1 dm"^3 = "0.010 M"`

The pH of the solution is defined as

`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 find

`"pH" = - log(0.010) = color(green)(|bar(ul(color(white)(a/a)2color(white)(a/a)|)))`