Question

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.

Answer 1

Answer:
`7.87* 10^(-3)M`

Step-by-step explanation:

`CH_3NH_2+H_2O\rightleftharpoons CH_3NH_3^++OH^-`

cM 0 0

`c-c\alpha`
`c\alpha`
`c\alpha`

So dissociation constant will be:

`K_b=((c\alpha)^(2))/(c-c\alpha)`

Give c= 0.125 M and
`\alpha` = ?

`K_b=5.25* 10^(-4)`

Putting in the values we get:

`5.25* 10^(-4)=((0.125* \alpha)^2)/((0.125-0.125* \alpha))`

`(\alpha)=0.063`

`[OH^-]=c* \alpha`

`[OH^-]=0.125* 0.063=7.87* 10^(-3)M`

Thus hydroxide ion concentration of a 0.125 molar aqueous solution of methylamine is
`7.87* 10^(-3)M`