Answer:
[tex]pH=11[/tex]
Explanation:
Hello there!
In this case, since the ionization of the described base is:
[tex]base+H_2O\rightarrow OH^-+baseH^+[/tex]
Thus, the equilibrium expression in terms of the reaction extent is:
[tex]Kb=5.86x10^{-6}=\frac{x^2}{0.187-x}[/tex]
Thus, by solving for x we obtain:
[tex]5.86x10^{-6}({0.187-x})=x^2\\\\1.10x10^{-6}-5.86x10^{-6}x-x^2=0[/tex]
So the values of x are:
[tex]x_1=0.0010M\\x_2=-0.0010M[/tex]
So the feasible answer is 0.0010 M, thus we compute the pOH:
[tex]pOH=-log(0.0010M)=3[/tex]
And therefore the pH:
[tex]pH=14-pOH=14-3\\\\pH=11[/tex]
Best regards!
