pOH = pKb + log [acid]/[base]
pH + pOH = 14
a)
0.500L x 0.900M NH4Cl = 0.450 mol
NH4+
0.500L x 0.900M NH3 = 0.450 NH3
0.450 + 0.100 = 0.550 mol NH4+
0.450 - 0.100 = 0.350 mol NH3
0.55 / 0.500 = 1.1M
0.35 / 0.500 =0.70M
Kb of NH3 = 1.8 x 10-5
pOH = 4.74 + log (1.1/.07)
pOH = 4.936
therefore;
pH = 14 -4.936
pH = 9.06
b)
Similarly;
0.500L x 0.800M NH4Cl = 0.400 mol
NH4+
0.500L x 0.200M NH3 = 0.100 NH3
0.400 + 0.100 = 0.500 mol NH4+
0.100 - 0.100 = 0 mol NH3
pOH =4.74; pH = 9.26
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