Question

Question #08b39

Answer 1

For part (b): `pH_"sol" = 9.13`

Explanation:

I will show you how to do part (b) so that you can solve part (a) on your own as practice.

So, you know that you're dealing with a buffer, which is a solution that contains a weak acid and its conjugate base, or a weak base and its conjugate acid.

For part (b), you have a buffer that contains ammonia, `"NH"_3`, a weak base, and ammonium chloride, `"NH"_4"Cl"`, the salt of ammonia's conjugate acid, the ammonium ion, `"NH"_4^(+)`.

Ammonium chloride is soluble in aqueous solution, so it will exist as ammonium cations and chloride anions, `"Cl"^(-)`.

The equilibrium reaction that describes the ionization of ammonia in aqueous solution looks like this

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

Since you have more conjugate acid than weak base, you can expect the pH of the solution to be a little smaller than it would have been for equal amounts of conjugate acid and weak base.

Make no mistake, the solution will still be quite basic, i.e. the pH will be bigger than 7, it's just that it will a little less basic.

By definition, the base dissociation constant, `K_b`, will be equal to

`K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3])`

You can rearrange this equation to solve for `["OH"^(-)]` at equilibrium

`["OH"^(-)] = K_b * (["NH"_3])/(["NH"_4^(+)])`

As you can see, when you have equal amounts of weak base and conjugate acid, `["OH"^(-)] = K_b`, which is equivalent to having `pOH = pK_b`.

Plug in your values to get

`["OH"^(-)] = 1.78 * 10^(-5) * (0.030color(red)(cancel(color(black)("M"))))/(0.040color(red)(cancel(color(black)("M")))) = 1.335 * 10^(-5)"M"`

This means that the solution's `pOH` will be

`pOH = - log(["OH"^(-)])`

`pOH = - log(1.335 * 10^(-5)) = 4.87`

Therefore, the solution's pH will be

`pH_"sol" = 14 - pOH = 14 - 4.87 = color(green)(9.13)`

For equal concentrations of ammonia and ammonium chloride, the pH would have been

`pH_"sol" = 14 - pK_b = 14 - 4.75 = 9.25`

Indeed, having more conjugate acid reduced the buffer's pH slightly.

Alternatively

You can use the Henderson-Hasselbalch equation to determine the `pOH` of the buffer. For a buffer that contains a weak base `"B"` and its conjugate cid `"BH"^(+)`, you have

`color(blue)(pOH = pK_b + log( (["BH"^(+)])/(["B"])))" "`

Here `pK_b` is equal to

`pK_b = - log(K_b)`

In your case, you have

`pOH = -log(1.78 * 10^(-5)) + log((0.040color(red)(cancel(color(black)("M"))))/(0.030color(red)(cancel(color(black)("M")))))`

`pOH = 4.75 + 0.12 = 4.87`

The pH will once again be

`pH_"sol" = 14 - 4.87 = 9.13`