50 ml of `0.1 M NH_(4)OH,` 25 ml of `2M NH_(4)Cl` were used to make a buffer. What is the pH if `pK_(a)` is 4.8?

Explanation

The given buffer is a base buffer.
For a base buffer `pOH=pK_(b)+log.(["Salt"])/(["Base"])`
`:. pOH=4.8+log {(((25xx2))/(75))/(((50xx0.1))/(75))}`
`=4.8+log(10)=4.8+1=5.8`
`: pH=(14.0pOH)=(14.0-5.8)=8.2`
`:. pH+pOH=14.0` at room temperature.