Answer : The
ion concentration is,
and the pH of a buffer is, 2.95
Explanation : Given,
![K_a=7.1\times 10^{-4}](https://tex.z-dn.net/?f=K_a%3D7.1%5Ctimes%2010%5E%7B-4%7D)
Concentration of
(weak acid)= 0.26 M
Concentration of
(conjugate base or salt)= 0.89 M
First we have to calculate the value of
.
The expression used for the calculation of
is,
![pK_a=-\log (K_a)](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%20%28K_a%29)
Now put the value of
in this expression, we get:
![pK_a=-\log (7.1\times 10^{-4})](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%20%287.1%5Ctimes%2010%5E%7B-4%7D%29)
![pK_a=4-\log (7.1)](https://tex.z-dn.net/?f=pK_a%3D4-%5Clog%20%287.1%29)
![pK_a=3.15](https://tex.z-dn.net/?f=pK_a%3D3.15)
Now we have to calculate the pH of the solution.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKNO_2%5D%7D%7B%5BHNO_2%5D%7D)
Now put all the given values in this expression, we get:
![pH=3.15+\log (\frac{0.89}{0.26})](https://tex.z-dn.net/?f=pH%3D3.15%2B%5Clog%20%28%5Cfrac%7B0.89%7D%7B0.26%7D%29)
![pH=2.95](https://tex.z-dn.net/?f=pH%3D2.95)
The pH of a buffer is, 2.95
Now we have to calculate the
ion concentration.
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![2.95=-\log [H_3O^+]](https://tex.z-dn.net/?f=2.95%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![[H_3O^+]=1.12\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1.12%5Ctimes%2010%5E%7B-3%7DM)
The
ion concentration is, ![1.12\times 10^{-3}M](https://tex.z-dn.net/?f=1.12%5Ctimes%2010%5E%7B-3%7DM)