Question

A buffer is prepared containing 0.75 M NH3 and 0.20 M NH4 . Calculate the pH of the buffer using the Kb for NH3. g

Answer 1

pH=8.676

Step-by-step explanation:

Given:

0.75 M
`NH_(3)`

0.20 M
`NH_(4)`

The objective is to calculate the pH of the buffer using the kb for
`NH_3`

Formula used:

`pOH=pka+log([salt])/([base])\\`

pH=14-pOH

Solution:

On substituting salt=0.75 and base=0.20 in the formula

`pOH=-log(1.77*10^(-5))+log(0.75)/(0.20)\\ =4.75+0.5740\\ =5.324`

pH=14-pOH

On substituting the pOH value in the above expression,

pH=14-5.324

Therefore,

pH=8.676