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
Find the pH of a buffer that consists of 0.82 M methylamin (CH3NH2) and 0.71 M CH3NH3Cl (pKb of methylamine (CH3NH2) = 3.35.)
I keep getting 3.41, but it says that it's wrong...
2 answers
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
CH3NH2=0.71M=B
CH3NH3Cl=0.82M=BH
pOH=pKb+log([BH]/[B]
pH=14-pOH
4 answers