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