Respuesta :
Answer : The value of pH of the solution is, 5.6
Explanation : Given,
The dissociation constant for methylamine = [tex]K_b=3.7\times 10^{-4}[/tex]
First we have to calculate the value of [tex]K_a[/tex].
[tex]K_a\times K_b=10^{-14}[/tex]
[tex]K_a\times (3.7\times 10^{-4})=10^{-14}[/tex]
[tex]K_a=2.7\times 10^{-11}[/tex]
The equilibrium chemical reaction will be:
[tex]CH_3NH_3^++H_2O\rightleftharpoons H_3O^++CH_3NH_2[/tex]
Initial conc. 0.2 0 0
At eqm. (0.2-x) x x
The expression for dissociation constant of acid is:
[tex]K_a=\frac{[H_3O^+][CH_3NH_2]}{[CH_3NH_3^+]}[/tex]
[tex]2.7\times 10^{-11}=\frac{(x)\times (x)}{(0.2-x)}[/tex]
By solving the term, we get the value of 'x'.
x = 2.3 × 10⁻⁶
x = -2.3 × 10⁻⁶
We are neglecting negative value of x.
Thus, the value of x = 2.3 × 10⁻⁶
[tex][H_3O^+]=x=2.3\times 10^{-6}M[/tex]
Now we have to calculate the value of pH.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (2.3\times 10^{-6})[/tex]
[tex]pH=5.6[/tex]
Therefore, the value of pH of the solution is, 5.6
