Answer:
pH=8.676
Explanation:
Given:
0.75 M [tex]NH_{3}[/tex]
0.20 M [tex]NH_{4}[/tex]
The objective is to calculate the pH of the buffer using the kb for [tex]NH_3[/tex]
Formula used:
[tex]pOH=pka+log\frac{[salt]}{[base]}\\[/tex]
pH=14-pOH
Solution:
On substituting salt=0.75 and base=0.20 in the formula
[tex]pOH=-log(1.77*10^{-5})+log\frac{0.75}{0.20}\\ =4.75+0.5740\\ =5.324[/tex]
pH=14-pOH
On substituting the pOH value in the above expression,
pH=14-5.324
Therefore,
pH=8.676
