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
Truong-Son N.
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)#

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