A 1.00 L buffer solution is 0.150 M in #HC_7H_5O_2# and 0.250 M in #LiC_7H_5O_2#. How do you calculate the pH of the solution after the addition of 100.0 mL of 1.00 M #HCl#? The Ka for #HC_7H_5O_2# is #6.5 × 10^–5#.

1 Answer
Apr 11, 2016

Answer:

You can do it like this:

Explanation:

It looks like the acid is benzoic acid #C_6H_5COOH#

For short - hand I'll call this #HA#

Benzoic acid dissociates:

#HA_((aq))rightleftharpoonsH_((aq))^(+)+A_((aq))^-" "color(red)((1))#

The lithium salt provides a large reserve of the co - base #A^-# by dissociating completely:

#LiA_((s))rarrLi_((aq))^(+)+A_((aq))^-" "color(red)((2))#

The initial no. moles of #A^-# is given by:

#nA_("init")^(-)=1xx0.25=0.25#

The no. #H^+# added is given by:

#nH^(+)=1xx100/1000=0.1#

From #color(red)((1))# you can see that these #H^+# ions will be absorbed by the large reserve of #A^-# ions shifting the equilibrium to the left thus "buffering" the pH.

You can see that they react in a 1:1 molar ratio so that the no. of moles of #A^-# remaining is given by:

#nA^(-)=0.25-0.1=0.15#

Each mole of #H^+# added will form 1 mole of #HA# so no. moles #HA# formed #=0.1#

The initial moles of #HA# is given by:

#nHA_("init") = 0.15xx1 =0.15#

So the total moles #HA# is given by:

#nHA=0.1+0.15=0.25#

The expression for the acid dissociation constant #K_a# is given by:

#K_a=([H_((aq))^(+)][A_((aq))^-])/([HA_((aq))])#

Rearranging:

#[H_((aq))^+]=K_axx([HA_((aq))])/([A_((aq))^-])#

I will ignore any changes in concentration due to the dissociation of the acid as it is very small.

The total volume is common to both acid and co - base and will cancel so we can write:

#[H_((aq))^+]=6.5xx10^(-5)xx0.25/0.15=1.083xx10^(-4)#

#pH=-log(1.083xx10^(-4))#

#color(red)(pH=3.96)#