Question

Question #26dfc

Answer 1

`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"`.

Answer 2

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]`