Answer:
[tex]\frac{[CH_3NH_3^+]}{[CH_3NH_2]}=4168.69[/tex]
Explanation:
As we know that:-
For [tex]CH_3NH_2[/tex]:-
[tex]pK_{b}=3.38[/tex]
pH = 7
Also, pH + pOH = 14
So, pOH = 14 - 7 = 7
Considering the Henderson- Hasselbalch equation for the calculation of the pOH of the basic buffer solution as:
[tex] pOH=pK_b+log\frac{[CH_3NH_3^+]}{[CH_3NH_2]} [/tex]
So,
[tex] 7=3.38+log\frac{[CH_3NH_3^+]}{[CH_3NH_2]} [/tex]
[tex]log\frac{[CH_3NH_3^+]}{[CH_3NH_2]} = 7-3.38=3.62[/tex]
[tex]\frac{[CH_3NH_3^+]}{[CH_3NH_2]}=4168.69[/tex]
