What are the pH of the solutions?

a). 0.10 mol of NaOH is added to 1.0 L of a solution that is 0.40 M CH3NH2 and 0.70 M CH3NH3Cl
b). 0.10 mol of HCl is added to 1.0 L of a solution that is 0.40 M CH3NH2 and 0.70 M CH3NH3Cl

1 Answer
Apr 13, 2018

They are both buffers. The starting "pH" is gotten via the Henderson-Hasselbalch equation.

"pH" = "pK"_a + log\frac(["B"])(["BH"^(+)])

You have not supplied the K_b for "CH"_3"NH"_2... it is your responsibility to look in your textbook for such a number. Good luck...

"pK"_a = -log(K_w/K_b) = ???

Adding "NaOH", it reacts with weak acid and makes weak base. Hence,

color(blue)("pH") = "pK"_a + log(("0.40 M B" cdot "1.0 L" + "0.10 mols NaOH")/("0.70 M BH"^(+) cdot "1.0 L" - "0.10 mols NaOH"))

= color(blue)("pK"_a - 0.079)

You can convince yourself that adding "0.10 mols HCl" means it reacts with weak base and forms weak acid, resulting in:

color(blue)("pH") = "pK"_a + log(("0.40 M B" cdot "1.0 L" - "0.10 mols HCl")/("0.70 M BH"^(+) cdot "1.0 L" + "0.10 mols HCl"))

= color(blue)("pK"_a - 0.426)