(a)
This is a weak acid:
sf(HOAcrightleftharpoonsH^++OAc^-)
This is a long question so I will go straight to this expression for the pH of a weak acid, which can be derived from the ICE table:
sf(pH=1/2[pK_a-loga])
sf(a) is the concentration of the acid.
sf(pK_a=-logK_a=-log[1.8xx10^(-5)]=4.745)
:.sf(pH=1/2[4.75-(-2)])
sf(pH=color(red)(3.37))
(b)
This is the salt of a weak acid and a strong base so undergoes hydrolysis:
sf(OAc^(-)+H_2OrightleftharpoonsHOAc+OH^-)
sf(pK_a+pK_b=pK_w=14)
:.sf(pK_b=14-4.75=9.255)
We can use the corresponding expression for the pOH of a weak base:
sf(pOH=1/2[pK_b-logb])
sf(b) is the concentration of the base.
sf(pK_a+pK_b=14)
:.sf(pK_b=14-4.475=9.255)
:.sf(pOH=1/2[9.255-(-2)]=5.627)
sf(pH+pOH=14)
:.sf(pH=14-5.627=color(red)8.37)
(c)
Carbonate ions are weak bases:
sf(CO_3^(2-)+H_2OrightleftharpoonsHCO_3^(-)+OH^-)
sf(K_(b1)xxK_(a2)=K_w=10^(-14))
sf(K_(b1)=10^(-14)/(5.6xx10^(-11))=1.786xx10^(-4))
sf(pK_(b1)=-log[K_(b1)]=-log[1.786xx10^(-4)]=4.748)
sf(pOH=1/2[pK_(b1)-logb])
sf(pOH=1/2[4.78-(-2)]=3.39)
sf(pH=14-3.39=color(red)(10.61)
(d)
Carbonic acid is a weak acid:
sf(H_2CO_3rightleftharpoonsH^(+)+HCO_3^-)
sf(pK_(a1)=-log[K_(a1)]=-log[4.3xx10^(-7)]=6.37)
If sf(K_(a1) is greater than sf(K_(a2) by a factor of sf(10^3) then we can assume that virtually all the sf(H^+) ions come from the 1st dissociation.
:.sf(pH=1/2[6.37-(-2)]=color(red)4.19)
(e)
sf(HCO_3^(-)rightleftharpoonsH^(+)+CO_3^(2-))
sf(K_(a2)=5.6xx10^(-11))
sf(pK_(a2)=-log[K_(a2)]=-log[5.6xx10^(-11)]=10.25)
sf(pH=1/2[10.25-(-2)]
sf(pH=color(red)(6.13)