Answer : The
pH of a solution is, 8.56
Explanation : Given,
![K_b=1.8\times 10^{-5}](https://tex.z-dn.net/?f=K_b%3D1.8%5Ctimes%2010%5E%7B-5%7D)
Concentration of ammonia (base) = 0.10 M
Concentration of ammonium nitrate (salt) = 0.55 M
First we have to calculate the value of
.
The expression used for the calculation of
is,
![pK_b=-\log (K_b)](https://tex.z-dn.net/?f=pK_b%3D-%5Clog%20%28K_b%29)
Now put the value of
in this expression, we get:
![pK_b=-\log (1.8\times 10^{-5})](https://tex.z-dn.net/?f=pK_b%3D-%5Clog%20%281.8%5Ctimes%2010%5E%7B-5%7D%29)
![pK_b=5-\log (1.8)](https://tex.z-dn.net/?f=pK_b%3D5-%5Clog%20%281.8%29)
![pK_b=4.7](https://tex.z-dn.net/?f=pK_b%3D4.7)
Now we have to calculate the pOH of buffer.
Using Henderson Hesselbach equation :
![pOH=pK_b+\log \frac{[Salt]}{[Base]}](https://tex.z-dn.net/?f=pOH%3DpK_b%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BBase%5D%7D)
Now put all the given values in this expression, we get:
![pOH=4.7+\log (\frac{0.55}{0.10})](https://tex.z-dn.net/?f=pOH%3D4.7%2B%5Clog%20%28%5Cfrac%7B0.55%7D%7B0.10%7D%29)
![pOH=5.44](https://tex.z-dn.net/?f=pOH%3D5.44)
The pOH of buffer is 5.44
Now we have to calculate the pH of a solution.
![pH+pOH=14\\\\pH+5.44=14\\\\pH=14-5.44\\\\pH=8.56](https://tex.z-dn.net/?f=pH%2BpOH%3D14%5C%5C%5C%5CpH%2B5.44%3D14%5C%5C%5C%5CpH%3D14-5.44%5C%5C%5C%5CpH%3D8.56)
Thus, the pH of a solution is, 8.56