Question

Question #eab13

Answer 1

`sf(pH=4.1)`

Explanation:

I got a different value for the initial pH. Here's what I got:

Benzoic acid is a weak acid and dissociates:

`sf(C_6H_5COOHrightleftharpoonsC_6H_5COO^(-)+H^+)`

For which:

`sf(K_a=([C_6H_5COO^-][H^+])/([C_6H_5COOH]))`

Rearranging:

`sf([H^+]=K_(a)xx([C_6H_5COOH])/([C_6H_5COO^-]))`

Since `sf(pK_a=4.2)` this means `sf(K_a=6.31xx10^(-5)color(white)(x)"mol/l")`

I will assume that the no. of moles given refers to equilibrium moles. Since the volume is 1L we can write:

`sf([H^+]=6.31xx10^(-5)xx0.3/0.35=5.4xx10^(-5)color(white)(x)"mol/l")`

`sf(pH=-log[H^+]=-log(5.4xx10^(-5))=4.26)`

Now a small amount of HCl is added. We have a buffer solution here with a large reserve of benzoate ions which can absorb the addition of small amounts of `sf(H^+)` ions :

`sf(C_6H_5COO^(-)+H^(+)rarrC_6H_5COOH)`

If we add 0.05 moles of `sf(H^+)` ions you can see that 0.05 moles of benzoate ions will be consumed and 0.05 moles of benzoic acid will be formed.

So we can say:

`sf(n_(C_6H_5COO^-)=0.35-0.05=0.3)`

and

`sf(n_(C_6H_5COOH)=0.3+0.05=0.35)`

There may be a volume change on mixing but this does not matter as the volume is common to acid and co - base so cancels.

We can now put these values into our original expression for `sf([H^+]rArr)`

`sf([H^+]=6.31xx10^(-5)xx0.35/0.3=7.36xx10^(-5)color(white)(x)"mol/l")`

`:.` `sf(pH=-log(7.36xx10^(-5))=4.1)`

You can see that the buffer has worked since the pH has fallen only very slightly.