d 01 M Na2CO3 3 Using the internet list at least 5 Ir that m

d. 0.1 M Na2CO3 3. Using the internet, list at least 5 Ir that may be used as pH i

To predict the pH of 0.1M Na2CO3,:

The base hydrolysis in solution takes place as follows:

Na2CO3 +H2O <--->2Na+ +HCO3- +OH-

CO3^2- +H2O <--->HCO3- +OH-

,kb1=kw/ka2=10^-14/(4.69*10^-11)=0.000213=2.13*10^-4

HCO3- +H2O <---> H2CO3 +OH-

kb2=kw/ka1=10^-14/(4.45*10^-7)=2.25*10^-8

kb1=[HCO3-][OH-]/[CO3^2- ]

2.13*10^-4=x^2/(0.1-x) [for very low dissociation ,0.1>>x]

2.13*10^-4=x^2/(0.1)

or, x=0.461*10^-2=[OH-]=[HCO3-]

Now,kb2=2.25*10^-8=[H2CO3][OH-]/[HCO3-]

ICE table

2.25*10^-8=[H2CO3][OH-]/[HCO3-]=y*(0.461*10^-2+y)/(0.461*10^-2-y)

or,0.461*10^-2<<<y for very low kb2 or dissociation

2.25*10^-8=y*(0.461*10^-2)/(0.461*10^-2)=y

y=2.25*10^-8=kb2=[OH-]

[H3O+]=kw/[OH-]=(10^-14)/(2.25*10^-8)=4.44*10^-7M

pH=-log[H3O+]=-log (4.44*10^-7M)=6.3

pH=6.3