Consider a solution of `CH_3COONH_4` which is a salt weak acid and weak base.
The equilibrium involved in the solutions are :
`CH_3COO^(-)+H_2OhArrCH_3COOH+OH^(-) " ........"(i)`
`NH_(4)^(+)+H_(2)hArrNH_(4)OH+H^(+)" ........"(ii)`
`H^(+)+OH^(-)hArrH_(2)O" ........"(iii)`
If we add these reactions, then the net reaction is :
`CH_(3)COO^(-)+H_(2)^(+)+H_(2)OhArrCH_(3)COOH+NH_(4)OH" ........"(iv)`
Both `CH_(3)COO^(-)`and `NH_(4)^(+)` get hydrolysed independently and their hydrolysis depends on :
(a) their initial concentration
(b) The value of `K_(h)` which is `(K_(w))/(K_(a))` for `CH_(3)COO^(-)` and `(K_(w))/(K_(b))` for `NH_(4)^(+)`.
Since both of the ions were produced form the salt, their initial concertration are same. Therefore, unless and until the value of `(K_(w))/(K_(a))` or `K_(a)` and `K_(b)` is same, the degree of hydrogen of ions can't be same.
To explain why we assume that degree of hydrolysis of cation and anion is same, we needed to now look at the third reaction i.e.,combination of `H^(+)` and `OH^(-)` ions. It is obvious that this reaction happens only because one reaction produced `H^(+)` ion and the other prodcued `OH^(-)` ions. We can also note that this reaction causes both the hydrolysis reaction to occur more since their product ions are being consumed. Keep this in mind that the equilibrium which has smaller value of the equilibrium constant is affected more by the common ion effect. For the same reason if for any reson a reaction is made to occur to a greater extent by the consumption of any of the prodcut ion, the reaction with the smaller value of equilibrium constant tends to get affected more.
Therefore, we conclude that firstly the hydroylsis of both the ions occurs more in the presence of each other (due to consumption of the product ions) than in each other's absence. Secondly,the hydroylsis of the ion which occurs to a lesser extent (due to smaller value of `K_(h))`is affected more than the one whole `K_(h)` is greater. Hence we can see that the degree of hydroylsis of both the ions would be close to each other when they are getting hyderolysed in the presence of each other.
In a solution of `NaHCO_(3)`, the amphroximately same undergo ionization to form `H_(+)` ion and hydrolysis to form `OH^(-)` ion.
`HCO_(3)^(-)+H_(2)Ooverset("ionization")hArrCO_(3)^(2-)+H_(2)O`
`HCO_(3)^(-)+H_(2)Ooverset("hydrolysis")hArrH_(3)CO_(2)+OH^(-)`
To calculate ph, suitable approximation is :

Options

A. `[CO_3^2-]=[H_2CO_3]`
B. degree of ionization = degree of hydrolysis
C. Both (A) and (B)
D. neither 'A' and 'B'

Answer

C

Explanation

`because pH=((PK_(a_1))+(PK_(a_2)))/2`