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Calculate the `pH` of a solution made by mixing `0.1 M NH_(3)` and `0.1M (NH_(4))_(2)SO_(4). (pK_(b)` of `NH_(3) = 4.76)` |
Answer» It is a basic buffer. `["Base"] = [NH_(3)] = 0.1m` `["Salt"] = [(NH_(4))_(2)SO_(4)] = 0.1 xx2 = 0.2M` `[(NH_(4))_(2) SO_(4) rarr 2overset(o+)NH_(4) +SO_(4)^(2-)]` `pOH = pK_(b) + "log"(["Salt"])/(["Base"])` `4.76 +log (0.2M)/(0.1M)` `= 4.76 +log2` `= 4.76 +0.3 = 5.03` `:. pH = 14 - 5.03 = 8.97` |
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