See H+ concentration will come from 2 reactions.
So, [H+]= root(K1×K2)
pH = - log(4×10^-9)
But why don't we apply salt hydrolysis here?
Its an amphiprotic acid, we have direct relation.
Also, this isn't a case of salt hydrolysis.
for salts of dibasic acids they tend to hydrolyse as well as release hydrogeen
so when you write equations for both eqb and neglect small terms you would get same result
I'm weak at ionic.....
NaHCO3 is a salt of strong base (NaOH) and weak acid acid (H2CO3) right?
Also since only HCO3- is attached to Na we will consider Ka1....
So by formula -
pH=7+1/2( pKa1+ log C)
Won't we use this?
as HCO3- can release a hydrogen or under go salt hydrolysis we cant use it
we can use it only for hydrolysing salt
@B_Rama_Chandra @Manan_Upadhyay can you please send notes for the above question, whenever you have time.
see this explained well.
Forgive my bad writing. Better see from the video. Method in notes is slights different ig.
Handwriting is pretty legible