Suppose 20.0 mL of a 0.100 M NH3 with 0.100 M HBr. Determine the pH of the solution after the addition of 5.00 mL and then 15.00 mL.
1)
mmoles of NH3 = 20.0 x 0.100 = 2.0
mmoles of HBr = 5.00 x 0.100 = 0.50
NH3 + HBr ---------------> NH4+
2.0 0.5 0
1.5 0 0.5
pH = pKa + log [base / acid]
= 9.26 + log [1.5 / 0.5]
pH = 9.73
2)
mmoles of HBr = 15 x 0.100 = 1.50
NH3 + HBr ---------------> NH4+
2.0 1.5 0
0.5 0 1.5
pH = pKa + log [base / acid]
= 9.26 + log [0.5 / 1.5]
pH = 8.78
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