Question

What is the pH of 0.1M NaHCO3? K1 = 4.5 x 10-7 , K2 = 4.5 x 10-11 for carbonic Acids?

Answer 1

NaHCO3 + H2O <---> H2CO3 + NaOH 

Kb= kw/ka  

Where ka is equal to K1=4.5 x 10^ -7  

So kb=(10^-14)/(4.5*10^-7) is equal to 2.22 x 10^-8 

Dissociation is low 2.22 x 10^8 = [OH] ^ 2/0.1 

[OH] is equal to 4.7 x 10 ^ -5  

POH = -log[OH] = 4.32 

So pH = 14-4.32=9.68