Question

Aweak acid HA, has a pKa of 5.0. If 1.0 mol of this acid and 0.1 mol of NaOH were dissolved in one liter of water, what would the final pH be?

Answer 1

`sf(pH=4.04)`

Explanation:

The weak acid is partly neutralised by the alkali:

`sf(HA_((aq))+NaOH_((aq))rarrNaA_((aq))+H_2O_((l)))`

We start with 1 mole of HA and add 0.1 moles of NaOH.

Since you can see that they react in a 1:1 molar ratio we can say that the number of moles of HA remaining is 1 - 0.1 = 0.9.

This means that the number of moles of NaA formed is 0.1.

Effectively we have created an acidic buffer which consists of a mixture of a weak acid and its co - base.

We can find the pH using the expression for `sf(K_a)`.

HA is a weak acid and dissociates:

`sf(HArightleftharpoonsH^++A^-)`

For which:

`sf(K_a=([H^(+)][A^-])/([HA])=10^(-pK_a)=10^(-5)color(white)(x)"mol/l"" "color(red)((1)))`

`:.` `sf([H^+]=K_axx([HA])/([A^-])" "color(red)((2)))`

Note that these are equilibrium concentrations.

To find the concentrations at equilibrium we can set up an ICE table:

`sf(" "HA" "rightleftharpoons" "H^+" " +" "A^-)`

`sf(color(red)(I)" "0.9" "0" " 0.1)`

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

`sf(color(red)(E)" "(0.9-x)" "x" "(0.1+x))`

`:.` `sf((x(0.1+x))/((0.9-x))=10^(-5)" "color(red)((3)))`

At this point I am going to make the assumption that, because the value of `sf(K_a)` is so small, the value of `sf(x)` is negligible compared to `sf(0.1)` and `sf(0.9)`.

So we can say:

`sf((0.1+x)rArr0.1)`

and

`sf((0.9-x)rArr0.9)`

So `sf(color(red)((3))` becomes:

`sf((0.1x)/(0.9)=10^(-5))`

`:.` `sf(x=10^(-5)xx0.9/0.1=9xx10^(-5)color(white)(x)"mol/l"=[H^+])`

`sf(pH=-log[H^+]=-log(9xx10^(-5))=4.04)`

This derivation is given to explain how to justify the assumptions that are made.

If the value of `sf(K_a)` is small, as is the case here, then the initial concentrations of `sf(HA)` and `sf(A^-)` are a close approximation to the equilibrium concentrations.

In practice, there is no need to set up an ICE table and you can go straight to equation `sf(color(red)((2))`.

You can use moles instead of concentrations as the volume is common to both acid and co - base so cancels.

However, it is a good thing to state the assumptions which I have pointed out as well as the fact that I have ignored the tiny amount of `sf(H^+)` ions which arise from the auto - ionisation of water.