When 10 mL of 0.1 M acetic acid (pKa = 5.0) is titrated against 10 mL of 0.1 M ammonia solution (pKb = 5.0),
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pKa=−logKw,pKb=−logKbpKa=−logKw,pKb=−logKb
pH=−12[logKa+logKw−logKb]pH=−12[logKa+logKw−logKb]
=−12[−5+log(1×10−14)−(−5)]=−12[−5+log(1×10−14)−(−5)]
=−12[−5−14+5]=−12(−14)=7
pH=−12[logKa+logKw−logKb]pH=−12[logKa+logKw−logKb]
=−12[−5+log(1×10−14)−(−5)]=−12[−5+log(1×10−14)−(−5)]
=−12[−5−14+5]=−12(−14)=7
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