(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)#
