#(a)#
Ethanoic acid is a weak acid and dissociates:
#CH_3COOH_((aq))rightleftharpoonsCH_3COO_((aq))^(-)+H_((aq))^+#
For which:
#K_a=([CH_3COO_((aq))^(-)][H_((aq))^+])/([CH_3COOH_((aq))])=1.8xx10^(-5)"mol/l"# at #25^@"C"#
Rearranging gives:
#[H_((aq))^+]=K_axx([CH_3COOH_((aq))])/([CH_3COO_((aq))^-])#
I am going to assume that equilibrium concentrations have been given so:
#[H_((aq))^+]=1.8xx10^(-5)xxcancel(0.25)/cancel(0.25)#
#pH=-log[H_((aq))^+]=-log(1.8xx10^(-5))#
#color(red)(pH=4.74)#
#(b)#
The buffer contains a large reserve of the co - base which can absorb the addition of a small amount of #H^+# ions:
#CH_3COO_((aq))^(-)+H_((aq))^(+)rarrCH_3COOH_((aq))#
The initial no. of moles of #CH_3COO_((aq))^-# is given by:
#n=cxxv=0.25xx250/1000=0.0625#
The no. of moles of #H^+# added we are told, is #0.005#.
From the equation we can see that they react in a 1:1 mole ratio.
So the no. moles #CH_3COO_((aq))^-# remaining is given by:
#n_(CH_3COO^-)=0.0625-0.005=0.0575#
The no. moles of #CH_3COOH# will have increased by the same amount.
The initial no. moles of #CH_3COOH# is given by:
#n=cxxv=0.25xx250/1000=0.0625#
So the total moles #CH_3COOH_((aq))# remaining is given by:
#n_(CH_3COOH)=0.0625+0.005=0.13#
We don't know the new total volume but since this is common to acid and co - base it will cancel so we can write:
#[H_((aq))^+]=1.8xx10^(-5)xx0.13/0.0575=4.069xx10^(-5)#
#pH=-log(4.069xx10^(-5))#
#color(red)(pH=4.39)#
#(c)#
If a small amount of #OH_((aq))^-# ions is added these will react with #CH_3COOH_((aq))#:
#CH_3COOH_((aq))+OH_((aq))^(-)rarrCH_3COO_((aq))^(-)+H_2O_((l))#
We have already calculated the initial moles of #CH_3COOH# to be #0.0625#.
You can see from the equation that they react with #OH^-# in a 1:1 molar ratio so the no. moles remaining is given by:
#n_(CH_3COOH)=0.0625-0.005=0.0575#
The no. moles of #CH_3COO^-# will have increased by the same amount.
#:.n_(CH_3COO^-)=0.0625+0.005=0.13#
Again, since the new total volume is constant we can write:
#[H_((aq))^+]=1.8xx10^(-5)xx0.0575/0.13#
#[H_((aq))^+]=0.796xx10^(-5)#
#color(red)(pH=5.09)#
