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Whatr is ph of 1L buffer containig initially 0.02 mol NH3 and 0.02 mol NH4Cl after addition of 0.01 mol of NaOH? (Ignore volume change).pKa NH4+=9.24
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5 Answers
Manish Shankar
·2009-01-11 09:48:25
I m also getting log3
pH = Pka +log [(NH3)/NH4+)
pH1=pKa
pH2=pKa+log(0.03/0.01)=9.24+lpg3=9.24+0.48=9.72