Posted by Edward on Sunday, March 3, 2013 at 2:36am.
CH3NH2=0.71M=B
CH3NH3Cl=0.82M=BH
B+H2O--->BH +OH
kb=[OH][BH]/{B}
kb=[OH]*([BH]/[B])
-logkb=-log[OH]*-log([BH]/[B])
pkb=pOH-log([BH]/[B])
pOH=pKb+log([BH]/[B]
pH=14-pOH
Essentially, I was thinking this problem out by deriving the Henderson-Hasselbalch equation. You only need the following information to calculate your answer:
CH3NH2=0.71M=B
CH3NH3Cl=0.82M=BH
pOH=pKb+log([BH]/[B]
pH=14-pOH
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