Question

Calculate the pH of a buffer solution containing 0.2 mole of NH4Cl and 0.1 mole of
NH4OH per litre. Given Kb for NH4OH = 1.85 X 10-5

Answer 1

According to Henderson – Hasselbalch equation:

pOH = pKb + log ([salt]/ [base])

pKb = – log Kb = – log .85 x 10-5 = 4.733

Therefore, pOH = 4.733 + log (0.2 / 0.1)

= 4.733 + 0.301 = 5.034

pH = 14 – pOH = 14 – 5.034 = 8.966

Answer 2

The pH of the solution is 8.97

Explanation:

The chemical equation for the reaction of ammonium hydroxide and hydrochloric acid follows:

$NH_4OH+HCl\rightarrow NH_4Cl+H_2O$

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[NH_4Cl]}{[NH_4OH]})$

We are given:

$pK_b$ = negative logarithm of base dissociation constant of ammonium hydroxide = 4.73

$[NH_4Cl]=\frac{0.2}{1}$

$[NH4OH]=\frac{0.1}{1}$

pOH = ?

Putting values in above equation, we get:

$pOH=4.73+\log(\frac{0.2/1}{0.1/1})\\\\pOH=5.03$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\\\pH=14-5.03=8.97$

Learn more about buffer solution:

https://brainly.com/question/13386397

https://brainly.com/question/13264722

#learnwithbrainly