Question

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`.

Answer 1

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)`