What is the change in pH after addition of 10.0 mL of 1.0 M sodium hydroxide to 90.0 mL of a 1.0 M NH3/1.0 M NH4+ buffer? [Kb for ammonia is 1.8 x 10-5]
pH of 90 mL of 1 M NH3/1 M NH4+ buffer = pKa + Log (nNH3/nNH4+)
Here, Kb = 1.8 x 10-5.
We know that: pKb = -Log(Kb)
= -Log(1.8 x 10-5)
= 4.74
We also know that: pKa + pKb = 14
i.e. pKa = 14 - 4.74
pKa = 9.26
And nNH3 = 1*90, i.e 90 mmol., nNH4+ = 90 mmol, i.e. nNH3/nNH4+ = 1
Therefore, pH = 9.26
No. of mmol. of NaOH added = 10*1, i.e. 10 mmol.
i.e. NH3 (90 mmol) + NH4+ (90 mmol) + NaOH (10 mmol) = NH3 (100 mmol) + NH4+ (80 mmol) + NaX (10 mmol)
pH = pKa + Log (nNH3/nNH4+)
= 9.26 + Log(100/80)
= 9.26 + 0.097
Therefore, the required change in pH = (9.26 + 0.097) - 9.26, i.e. 0.097 ~ 0.1.
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