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

Answer:

pH=8.676

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:

$pOH=pka+log\frac{[salt]}{[base]}$

pH=14-pOH

Solution:

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

$pOH=-log(1.77*10^{-5})+log\frac{0.75}{0.20}\\ =4.75+0.5740\\ =5.324$

pH=14-pOH

On substituting the pOH value in the above expression,

pH=14-5.324

Therefore,

pH=8.676