Question

The pH of 0.2 M aqueous solution of NH4Cl will be (pKb of NH3 = 4.74; log 2 = 0.3)

Answer 1

Quantity of the solution = 0.2M (Given)

pKb of the solution = 4.74  (Given)

Quantity of salt = 0.3M

Quantity of base = 0.2M

Thus,

pOH = pKb + log ( salt/base)

Substitute into equation:

pOH = 4.74 + log (0.3/0.2)

pOH = 4.74 + log 2

pOH = 4.74 + 0.30

pOH = 5.04

Total ph will be -

pH = 14.00- pOH

pH = 14.00-5.04

pH = 8.96

Therefore, the pH of the solution will be 8.96.

Answer 2

To find pH of the the solution, we can use the following formula:

pH + pOH = 14

First let's solve for pOH.

given data:

Molarity of the solution: 0.2 M

pKb of NH3= 4.74 , log2 = 0.3

pOH = pKb + log (salt/base)

putting in values,

pOH = 4.74 + log (0.3/0.2)

pOH = 4.74 + log 2 = 4.74 + 0.3

pOH = 5.04

So, now pH will be

pH + 5.04 = 14

or,

pH = 14 - 5.04

pH = 8.96

is the required pH of the solution.

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