initially only NH3 present.
for weak acid
pOH = 1/2 [pKb - log C]
pKb = - log Kb = - log [1.75 x 10-5] = 4.76
pOH = 1/2 [4.76 - log 0.250]
pOH = 2.68
pH = 14 - 2.68
pH = 11.32
2) initially
millimoles of NH3 = 5.00 x 0.250 = 1.250
millimoles of HCl added = 5.00 x 0.100 = 0.50
after HCl added
millimoles of NH3 left = 1.250 - 0.50 =0.750
millimoles of NH4+ = 0.50
total volume = 5.00 + 5.00 = 10.00 mL
[NH3] = 0.750 / 10.00 = 0.075 M
[NH4+] = 0.50 / 10.00 = 0.05 M
pOH = pKb + log [NH4+] / [NH3]
pOH = 4.76 + log [0.05] / [0.075]
pOH = 4.58
pH = 14 - 4.58
pH = 9.42
5.00 mL of 0.250 M ammonia (NH3) is titrated with 0.100 M hydrochloric acid (HCl). The...
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