Calculate the pH of a solution prepared by mixing equal volumes of 0.23M methylamine (CH3NH2, Kb = 3.7×10−4) and 0.64M CH3NH3Cl.?

The most common form of the Henderson-Hasselbalch equation is this:

pH = pKa + log (base/acid)

I have a brief discussion here:

http://chemteam.info/AcidBase/HH-Equation.html

You can write an alternate form of the HH using pOH and pKb:

pOH = pKb + log (acid/base)

I'll use the alternate form for this problem.

pOH = 3.4318 + log (0.64/0.23) <--- comment below about the ratio

pOH = 3.4318 + 0.4444 = 3.8762

pH = 14.0000 - 3.8762 = 10.1238

To two sig figs, this is 10.12

Why did I use 0.64/0.23? After all, the 0.23 and the 0.64 get diluted by the mixing.

Yes, they do get diluted but the ration stays the same. By using equal volumes, each molarity will get cut in half, so let's try that new ratio:

0.32 / 0.115 = 2.7826

log 2.7826 = 0.4444

You can also think of the ratio this way:

(0.64/2) / (0.23/2)

and the twos in the denominators of each fraction cancel out.

Here's a problem like yours:

http://www.jiskha.com/display.cgi?id=1362296196

Note that the question asker got 3.41 and didn't realize that that value was the pOH and needed to be converted to the pH by 14 minus the pOH for an answer of 10.59