What should be the concentration of NaA, if its 50 mL solution of `0.10MNH_(3)` and `0.10MNH_(4)CI` without changing the pH by more than `1.0` unit ? Assume no change in volume. `(K_(a) for HA= 1.0xx10^(-5))`

Explanation

Let a be the concentration of NaA.
After mixing, `M_(NaA)=(axx50)/(100)=(a)/(2)`
and `M_(HA)=(0.2xx50)/(100)=0.1`
Now `pH= -log K_(a)+log((["Salt"])/(["Acid"]))`
`4= -1.0xx10^(-5)+log ((a//2)/(0.1))`
`a=0.2M`