Methylamine, CH3NH2, is a weak base that reacts according to the reaction CH3NH2 + H2O <--> CH3NH3+ + OH- The value of the ionization constant, Kb, is 5.25 x 10 –4. Methylamine reacts to form salts such as methylammonium nitrate, (CH3NH3+)(NO3-). a. Calculate the hydroxide ion concentration, [OH-] of a 0.125 molar aqueous solution of methylamine.

Respuesta :

Answer:  [tex]7.87\times 10^{-3}M[/tex]

Explanation:

[tex]CH_3NH_2+H_2O\rightleftharpoons CH_3NH_3^++OH^-[/tex]

   cM                           0             0

 [tex]c-c\alpha[/tex]                      [tex]c\alpha[/tex]          [tex]c\alpha[/tex]

So dissociation constant will be:

[tex]K_b=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]

Give c= 0.125 M and [tex]\alpha[/tex] = ?

[tex]K_b=5.25\times 10^{-4}[/tex]

Putting in the values we get:

[tex]5.25\times 10^{-4}=\frac{(0.125\times \alpha)^2}{(0.125-0.125\times \alpha)}[/tex]

[tex](\alpha)=0.063[/tex]

[tex][OH^-]=c\times \alpha[/tex]

[tex][OH^-]=0.125\times 0.063=7.87\times 10^{-3}M[/tex]

Thus hydroxide ion concentration of a 0.125 molar aqueous solution of methylamine is [tex]7.87\times 10^{-3}M[/tex]