What is the pH of a solution prepared by dissolving 8.40 g of aniline hydrochloride C6H5NH3Cl in 750 mL of 0.210 M aniline,C6H5NH2?

Question

i kno to use the ph = pka + log (base/acid). getting pka isnt the issue, but what do i put into the log (base/acid)? (0.210/????)

Answer 1

The (base) = 0.210 M aniline.
The (acid) = 8.40 g/molar mass aniline hydrochloride (gives you moles) and that divided by 0.750 L = M

Answer 2

i keep getting docked off. im getting 5.00 but its wrong? the pka is 4.61978 and adding it to .38 for 5. -.-

Answer 3

Using your pKa I get 5.00 also. The only thing I can suggest is that you look up pKa again. My text gives a slightly different number of 4.59 which is enough to make pH = 4.97 or so.