Question #26dfc

2 Answers
Apr 27, 2017

Answer:

#C.#

Explanation:

We use the buffer equation where,

#pH=pK_a+log_(10){([H_3CNH_2])/([H_3CNH_3^+])}#

But we were quoted #pK_b# for #"methylamine"# NOT #pK_a# for #"methyl ammonium chloride"#.

Since in water, #pK_a+pK_b=14#, clearly, however, #pK_a=14-3.32=10.68#. Thus the concentration of #"methylamine"# is greater than that of #"methylamine hydrochloride"#.

Apr 27, 2017

Answer:

Excellent job. I just want to add that the same conclusion can be drawn using the pOH version of the Henderson - Hasselbalch Equation. pOH = pKb + #log([Conj Acid]/[Wk Base])#. It's just another approach in the mix.

Explanation:

Using:
pOH = pKb + #log_10([Conj Acid]/[Wk Base])#
Conj Acid => #CH_3NH_3^+#
Wk Base => #CH_3NH_2#

2.8 = 3.36 + #log_10([CH_3NH_3^+]/[CH_3NH_2])#
#log_10([CH_3NH_3^+]/[CH_3NH_2])# = 2.8 - 3.36 = -0.56
#([CH_3NH_3^+]/[CH_3NH_2])# = #10^(-0.56)# = 0.275
=> #[CH_3NH_3^+]:[CH_3NH_2] = # 0.275 : 1.00

=> #[CH_3NH_3^+] < [CH_3NH_2]#