Question

If a buffer solution is prepared from 5.15g `NH_4NO_3` and 0.10L of 0.15M `NH_3`, what is the pH of the solution?

Answer 1

`pH=pK_a+log_10{[[A^-]]/[[HA]]}`

Explanation:

It can be shown relatively easily that the `pH` of a weak acid mixed with its conjugate base in appreciable quantities conforms to the so-called `"buffer equation"` as given.

You have not quoted `pK_a` for ammonium ion, but I will take pity on you, because I know that `pK_a`, `NH_4^+ = 9.24`. Note that you are not expected to know these `pK_a` values from memory even as an undergraduate. You ARE expected to know that ammonium ion is a fairly weak acid, and how to manipulate the buffer equation.

Given the `pK_a` it is simply a question of determining the concentrations and plugging in the numbers:

`pH=9.24+log_10{[0.015*mol*L^-1]/[(5.15*g)/(80.04*g/(mol))xx1/(0.10*L)}}` `=` `??`

Note that here I knew that volume of the solution was `0.10*L`. How did I know this?

PS I hope you can see the units in my calculations, because I can't! It is important that the dimensions in the log argument cancel out, which I think they do!

Answer 2

`"pH=8.62"`

Explanation:

The ammonium ion is a weak acid and dissociates:

`NH_(4(aq))^+rightleftharpoonsNH_(3(aq))+H_((aq))^+`

For which:

`K_a=([NH_(3(aq))][H_((aq))^+])/([NH_(4(aq))^+])=5.6xx10^(-10)" ""mol/l"" "` at `25^@"C"`

To find the `"pH"` we need to calculate the concentration of the `H^+` ions.

Rearranging gives:

`[H_((aq))^+]=K_axx([NH_(4(aq))^+])/([NH_(3(aq))])" "color(red)((1))`

These are all equilibrium concentrations. We can set up an #"ICE" # table to find these.

For the initial moles of `NH_3` we use `n=cxxv`:

`:.nNH_(3(aq))=0.15xx0.1=0.015`

The initial moles of `NH_4NO_3` is equal to `m/M_r=5.15/80.04=0.0643`.

From the formula we can see that the initial moles of `NH_(4(aq))^+` must also be `0.0643`.

Now we can set up the `"ICE"` table:

`" "NH_4^+" "rightleftharpoons" "NH_3" "+" "H^+`

`color(red)"I"" "0.0643" "0.015" "0`

`color(red)("C")" "-x" "+x" "+x`

`color(red)("E")" "(0.0643-x)" "(0.015+x)" "x`

Because the value of `K_a` is so small the value of `x` is much smaller than `0.0643` and `0.015` so we can assume that:

`(0.0643-x)rArr0.0643`

and

`(0.015+x)rArr0.015`

Now we can put the numbers into `color(red)((1))rArr`

`[H_((aq))^+]=5.6xx10^(-10)xx0.0643/0.015=2.4xx10^(-9)" ""mol/l"`

You will note I can use moles and not concentrations as the volume is common to both `[NH_4^+]` and `[NH_3]` so cancels.

This also accounts for any volume change when the solid salt is added to the solution.

`pH=-log[H_((aq))^+]`

`:.pH=-log[2.4xx10^(-9)]=color(red)(8.62)`

In practice you could go straight to `color(red)((1))` but bear in mind the assumptions you have made in approximating equilibrium concentrations to initial ones.

This only works if `K_a` is very small. If not, then you need to use an `"ICE"` table.