Question

Question #48e56

Answer 1

Here's what's going on.

Explanation:

Your buffer consists of ammonia, a weak base, and ammonium chloride, the salt of its conjugate acid.

The species that are of interest here are ammonia, `"NH"_3`, and the ammonium ion, `"NH"_4^(+)`, which like I said is the conjugate acid of ammonia.

When you add hydrochloric acid, a strong acid, to the buffer, it will react with the ammonia to form more ammonium ions. The strong acid will be consumed in the process, provided that you have more ammonia present.

The reaction that's of interest to you will thus be

`"NH"_text(3(aq]) + "HCl"_text((aq]) -> "NH"_text(4(aq])^(+) + "Cl"_text((aq])^(-)`

Notice that you have `1:1` mole ratios across the board, This means that every mole of hydrochloric acid will consume 1 mole of ammonia and produce 1 mole of ammonium ions.

Your buffer contains

`C = n/V implies n = C * V`

`n_"ammonia" = "0.28 M" * "0.5 L" = "0.14 moles NH"""_3`

and

`n_"ammonium" = "0.22 M" * "0.5 L" = "0.11 moles NH"""_4^(+)`

After the reaction, you will be left with

`n_"ammonia" = 0.14 - 0.020 = "0.12 moles NH"""_3`

because the number of moles of ammonia decrease after reacting with the hydrochloric acid, and

`n_"ammonium" = 0.11 + 0.020 = "0.13 moles NH"""_4^(+)`

because ammonium ions are being produced by said reaction.

Remember that hydrochloric acid is completely consumed in the process.

Assuming that the volume of the buffer does not change (although this is very unlikely), the new concentrations of the weak base and conjugate acid will be

`["NH"_3] = "0.12 moles"/"0.5 L" = "0.24 M"`

and

`["NH"_4^(+)] = "0.13 moles"/"0.5 L" = "0.26 M"`

Now use the henderson-Hasselbalch equation to find the solution's `pOH`

`pOH = pK_b + log((["conjugate acid"])/(["weak base"]))`

`pOH = -log(1.8 * 10^(-5)) + log(( 0.26color(red)(cancel(color(black)("M"))))/(0.24color(red)(cancel(color(black)("M")))))`

`pOH = 4.74 + 0.035 = 4.78`

The pH of the solution will thus be

`pH_"sol" = 14 - pOH = 14 - 4.78 = color(green)(9.22)`