Question

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...

Answer 1

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

Answer 2

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