Question #eab13

1 Answer
Aug 16, 2016

Answer:

#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.