Question

Question #2c896

Answer 1

`["H"_3"O"^(+)] = 1.7 * 10^(-5)"M"`

Explanation:

Ammonium chloride, `"NH"_4"Cl"`, will dissociate completely in aqueous solution to produce ammonium cations, `"NH"_4^(+)`, and chloride anions, `"Cl"^(-)`

`"NH"_4"Cl"_text((aq]) -> "NH"_text(4(aq])^(+) + "Cl"_text((aq])^(-)`

Notice that the salt dissociates in a `1:1` mole ratio with the ammonium ion. This means that you get one mole of the latter for every mole of the former that is dissolved in solution.

So, if `0.20` moles of ammonium chloride are used to make this solution, it follows that the molarity of the ammonium ions will be

`n_(NH_4^(+)) = n_(NH_4Cl) = "0.20 moles"`

`c = "0.20 moles"/(400.0 * 10^(-3)"L") = "0.50 M"`

Once in aqueous solution, the ammonium ions will act as a weak acid and react with the water molecules to form ammonia, `"NH"_3`, a weak base, and hydronium ions, `"H"_3"O"^(+)`.

To determine the equilibrium concentration of the hydronium ions, use an ICE table

`" " "NH"_text(4(aq])^(+) + "H"_2"O"_text((l]) " "rightleftharpoons" " "NH"_text(3(aq]) " "+" " "H"_3"O"_text((aq])^(+)`

`color(purple)("I")" " " "0.50" " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " "0`
`color(purple)("C")" "color(white)(x)(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)`
`color(purple)("E")" "0.50-x" " " " " " " " " " " " " " " "x" " " " " " " " " " " " "x`

Now, notice that the poroblem provides you with the base dissociation constant, `K_b`, of ammonia. This means that you're going to have to use the equation

`color(blue)(K_a * K_b = K_W)" "`, where

`K_W` - the self-ionization constant of water, equal to `10^(-14)` at room temperature

to find the acid dissociation constant, `K_a`, of the ammonium ion.

`K_a * K_b = K_W implies K_a = K_W/K_a`

`K_a = 10^(-14)/(1.8 * 10^(-5)) = 5.6 * 10^(-10)`

By definition, the acid dissociation constant is equal to

`K_a = (["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])`

In your case, this will be equal to

`K_a = (x * x)/(0.50 - x) = x^2/(0.50 - x)`

Since the value of `K_a` is so small, you can use the approximation

`0.50 - x ~~ 0.50`

to get

`K_a = x^2/0.50 = 5.6 * 10^(-10)`

This means that you have

`x = sqrt(0.50 * 5.6 * 10^(-10)) = 1.7 * 10^(-5)`

Since `x` represents the equilibrium concentrations of ammonia and hydronium ions, it follows that you have

`["H"_3"O"^(+)] = color(green)(1.7 * 10^(-5)"M")`