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=pKa+log[B][BH+]

You have not supplied the Kb for CH3NH2... it is your responsibility to look in your textbook for such a number. Good luck...

pKa=−log(KwKb)=???

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

pH=pKa+log(0.40 M B⋅1.0 L+0.10 mols NaOH0.70 M BH+⋅1.0 L−0.10 mols NaOH)

=pKa−0.079

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

pH=pKa+log(0.40 M B⋅1.0 L−0.10 mols HCl0.70 M BH+⋅1.0 L+0.10 mols HCl)

=pKa−0.426

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