Respuesta :
Answer : The pH of the solution is, 9.18
Solution : Given,
Concentration (c) = 0.025 M
Base dissociation constant = [tex]k_b=9.1\times 10^{-9}[/tex]
The equilibrium reaction for dissociation of [tex]HONH_2[/tex] (weak base) is,
[tex]HONH_2+H_2O\rightleftharpoons HONH_3^++OH^-[/tex]
initially conc. c 0 0
At eqm. [tex]c(1-\alpha)[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
First we have to calculate the concentration of value of dissociation constant [tex](\alpha)[/tex].
Formula used :
[tex]k_b=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}[/tex]
Now put all the given values in this formula ,we get the value of dissociation constant [tex](\alpha)[/tex].
[tex]9.1\times 10^{-9}=\frac{(0.025\alpha)(0.025\alpha)}{0.025(1-\alpha)}[/tex]
By solving the terms, we get
[tex]\alpha=0.000603[/tex]
Now we have to calculate the concentration of hydroxide ion.
[tex][OH^-]=c\alpha=0.025\times 0.000603=1.5\times 10^{-5}M[/tex]
Now we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (1.5\times 10^{-5})[/tex]
[tex]pOH=4.82[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.82\\\\pH=9.18[/tex]
Therefore, the pH of the solution is, 9.18
