Given, Ka = 1.2 * 10-5
pKa = - Log (1.2 * 10-5)
pKa = 4.92
[Acid] = 0.210 M
[Conjugae base] = 0.540 M
According Henderson's and Hessalbatch's equation of buffers,
pH = pKa + Log[conjugate base] / [acid]
pH = 4.92 + Log(0.540/0.210)
pH = 5.33
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