Answer:
[tex][OH^-]=9.24x10^{-3}M[/tex].
[tex]pH=11.97[/tex].
Explanation:
Hello,
In this case, since the ionization of methylamine is:
[tex]CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)[/tex]
The equilibrium expression is:
[tex]Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}[/tex]
And in terms of the reaction extent [tex]x[/tex] which is equal to the concentration of OH⁻ as well as that of CH₃NH₃⁺ via ice procedure we can write:
[tex]3.7x10^{-4}=\frac{x*x}{024-x}[/tex]
Whose solution for [tex]x[/tex] via quadratic equation is 9.24x10⁻³ M since the other solution is negative so it is avoided. Therefore, the concentration of OH⁻ is:
[tex][OH^-]=x=9.24x10^{-3}M[/tex]
With which we can compute the pOH at first:
[tex]pOH=-log([OH^-])=-log(9.24x10^{-3})=2.034[/tex]
Then, since pH and pOH are related via:
[tex]pH+pOH=14[/tex]
The pH turns out:
[tex]pH=14-pOH=14-2.034\\\\pH=11.97[/tex]
Best regards.
