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?

1 Answer
Sep 5, 2016

Answer:

#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.