Question

use a table of ka and kb to determine pH of the 0.10 M CH3NH3Cl

d. 0.10 M CHÁNH,CI 4. Use the table of K and K values to predict whether a solution of NH CN is acidic, basic, or neutral. Explain your choice.

Answer 1

CH₃NH₃Cl is a salt weak base CH₃NH₂ and strong acid HCl.

CH₃NH₃Cl + H₂O  $\rightleftharpoons$ CH₃NH₂ + HCl

CH₃NH₃⁺ + H₂O $\rightleftharpoons$ CH₃NH₂ + H⁺

ICE table is

	CH3NH3-	CH3NH2	H+
I	C	-	-
C	-C$\alpha$	+C$\alpha$	+C$\alpha$
E	C - C$\alpha$	C$\alpha$	C$\alpha$

Dissociation constant is K_h , degree of dissociation is $\alpha$

kn = [CH3 NH2] [H+] [CH3 NH3t] [CH3 NH2] [H* ] [ok] [elg NH3] [own] on, Kn= Kw I or kh= - lid Kb

K, — ск к се — — к 1-2 ²1 kn = caz a is very less, hence ... K, - ск- or, x= Skole

LºnJ Kw = [H+] [OH-] K6 = [CH3 NH 3t] [of] [CH3NH2] l [ht] = ca ex J Khile гн*] - (x on, - 108 [H+] = - Rog (Ukaxe) - 24 ( Ku : ) Kb

Or, - log[H⁺] = - $\frac{1}{2}$ log Kw + $\frac{1}{2}$ logKb - $\frac{1}{2}$ logC

pH = $\frac{1}{2}$pKw - $\frac{1}{2}$pKb - $\frac{1}{2}$logC

Or, pH = 7 - $\frac{1}{2}$ (3.36 ) - $\frac{1}{2}$ log(0.10)

Or, pH = 7 - 1.68 + 0.5

Or, pH = 5.82

[ K_b of CH₃NH₂ = 4.36×10^-4, or pKb = 3.36]

4.

NH₄CN is salt of weak acid and weak base

Hence,

pH = 7 + $\frac{1}{2}$ (pKa - pKb)

Ka of HCN = 4.9×10^-10, pKb = - logKb = - log(4.9×10^-10) = 9.3

Kb of NH₃ = 1.8×10^-5 . Or, pKa = 4.75

Hence, pH = 7 + $\frac{1}{2}$ ( 9.30 - 4.75) = 7 + 2.275 = 9.275