Question

What is the pH change when 23.1 mL of 0.136 M is added to 74.9 mL of a buffer solution consisting of 0.148 M and 0.196 M ? ( for ammonium ion is .) pH change = _

Answer 1

Initial pH of the buffer is :

pH = pKa + log [base]/[acid]

pH = 9.252 + log (0.148)/(0.196)

pH = 9.1298

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No.of moles of base added = 23.1 mL x 0.136 M = 3.1416 m.mol

No.of moles of ammonium ions = 74.9 mL x 0.148 M = 11.0852 m.mol

No.of moles of base in buffer = 74.9 mL x 0.196 M = 14.6804 m.mol

[NH₃] = (11.0852 + 3.1416) / (74.9 mL + 23.1 mL ) = 0.14517 M

[NH⁴⁺] = (14.6804 - 3.1416) / (74.9 mL + 23.1 mL ) = 0.11774 M

Hence, Final pH will be:

pH = pKa + log (0.14517) / (0.11774)

pH = 9.252 + log (0.14517) / (0.11774)

pH = 9.342758

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Therefore, Change in pH:

Change in pH = 9.342758 - 9.1298

• Change in pH = 0.213