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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

Sagot :

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

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