Textbook 6D.13

[Barbara Soliman 1G]

Could someone explain how to do this problem for me? Specifically, how do you do solve for pH for part b? I got a different ranking than the solutions so I am not completely sure if I am doing this problem right. Thanks!

[Shria G 2D]

Hi,
For part b, the solution is 0.20 M CH3NH3Cl. Cl- doesn't affect the pH (as mentioned in last week's lecture) so it doesn't matter in the calculations. Table 6C.1 and 6C.2 in the textbook have acidity and basicity constants. CH3NH3 isn't there, but table 6C.2 says Kb for CH3NH2 is 3.6x10^-4. CH3NH2 is the conjugate base of the acid CH3NH3 because when CH3NH3 loses a proton, it becomes CH3NH2. Therefore, we can find the Ka for CH3NH3 by using the formula Ka x Kb = 10^-14. So, Ka = 10^-14/(3.6x10^-4) which equals 2.78x10^-11. Then you can set up an ICE table to find that the equilibrium constant is x^2/(0.2-x). If you set that equal to the Ka we found, you can solve for x which equals [H3O+]. If you take the -log of that, you will find pH. At the end, I got that pH = 5.63. I hope that helps!