Answer : The pH of the solution is, 9.18
Solution : Given,
Concentration (c) = 0.025 M
Base dissociation constant = ![k_b=9.1\times 10^{-9}](https://tex.z-dn.net/?f=k_b%3D9.1%5Ctimes%2010%5E%7B-9%7D)
The equilibrium reaction for dissociation of
(weak base) is,
![HONH_2+H_2O\rightleftharpoons HONH_3^++OH^-](https://tex.z-dn.net/?f=HONH_2%2BH_2O%5Crightleftharpoons%20HONH_3%5E%2B%2BOH%5E-)
initially conc. c 0 0
At eqm.
![c\alpha](https://tex.z-dn.net/?f=c%5Calpha)
First we have to calculate the concentration of value of dissociation constant
.
Formula used :
![k_b=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}](https://tex.z-dn.net/?f=k_b%3D%5Cfrac%7B%28c%5Calpha%29%28c%5Calpha%29%7D%7Bc%281-%5Calpha%29%7D)
Now put all the given values in this formula ,we get the value of dissociation constant
.
![9.1\times 10^{-9}=\frac{(0.025\alpha)(0.025\alpha)}{0.025(1-\alpha)}](https://tex.z-dn.net/?f=9.1%5Ctimes%2010%5E%7B-9%7D%3D%5Cfrac%7B%280.025%5Calpha%29%280.025%5Calpha%29%7D%7B0.025%281-%5Calpha%29%7D)
By solving the terms, we get
![\alpha=0.000603](https://tex.z-dn.net/?f=%5Calpha%3D0.000603)
Now we have to calculate the concentration of hydroxide ion.
![[OH^-]=c\alpha=0.025\times 0.000603=1.5\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3Dc%5Calpha%3D0.025%5Ctimes%200.000603%3D1.5%5Ctimes%2010%5E%7B-5%7DM)
Now we have to calculate the pOH.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
![pOH=-\log (1.5\times 10^{-5})](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%281.5%5Ctimes%2010%5E%7B-5%7D%29)
![pOH=4.82](https://tex.z-dn.net/?f=pOH%3D4.82)
Now we have to calculate the pH.
![pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.82\\\\pH=9.18](https://tex.z-dn.net/?f=pH%2BpOH%3D14%5C%5C%5C%5CpH%3D14-pOH%5C%5C%5C%5CpH%3D14-4.82%5C%5C%5C%5CpH%3D9.18)
Therefore, the pH of the solution is, 9.18