<u>Explanation:</u>
pH is the negative logarithm of hydronium ion concentration present in a solution.
- If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9
- If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
- The solution having pH equal to 7 is termed as neutral solution.
To calculate the pH of the solution, we use equation:
......(1)
To calculate the pOH of the solution, we use the equation:
pH + pOH = 14 ........(2)
We are given:
pH = 5.54
Putting values in equation 1, we get:
![5.54=-\log[H_3O^+]](https://tex.z-dn.net/?f=5.54%3D-%5Clog%5BH_3O%5E%2B%5D)
![[H_3O^+]=2.88\times 10^{-6}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D2.88%5Ctimes%2010%5E%7B-6%7DM)
Now, putting values in equation 2, we get:
14 = 5.54 + pOH
pOH = 8.46
The solution is acidic in nature.
We are given:
pOH = 9.7
Putting values in equation 2, we get:
14 = 9.7 + pH
pH = 4.3
Now, putting values in equation 1, we get:
![4.3=-\log[H_3O^+]](https://tex.z-dn.net/?f=4.3%3D-%5Clog%5BH_3O%5E%2B%5D)
![[H_3O^+]=5.012\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D5.012%5Ctimes%2010%5E%7B-5%7DM)
The solution is acidic in nature.
We are given:
pH = 7.0
Putting values in equation 1, we get:
![7.0=-\log[H_3O^+]](https://tex.z-dn.net/?f=7.0%3D-%5Clog%5BH_3O%5E%2B%5D)
![[H_3O^+]=1.00\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1.00%5Ctimes%2010%5E%7B-7%7DM)
Now, putting values in equation 2, we get:
14 = 7.0 + pOH
pOH = 7.0
The solution is neither acidic nor basic in nature.
We are given:
pH = 12.9
Putting values in equation 1, we get:
![12.9=-\log[H_3O^+]](https://tex.z-dn.net/?f=12.9%3D-%5Clog%5BH_3O%5E%2B%5D)
![[H_3O^+]=1.26\times 10^{-13}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1.26%5Ctimes%2010%5E%7B-13%7DM)
Now, putting values in equation 2, we get:
14 = 12.9 + pOH
pOH = 1.1
The solution is basic in nature.
We are given:
pOH = 1.2
Putting values in equation 2, we get:
14 = 1.2 + pH
pH = 12.8
Now, putting values in equation 1, we get:
![12.8=-\log[H_3O^+]](https://tex.z-dn.net/?f=12.8%3D-%5Clog%5BH_3O%5E%2B%5D)
![[H_3O^+]=1.58\times 10^{-13}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1.58%5Ctimes%2010%5E%7B-13%7DM)
The solution is basic in nature.
We are given:
![[H_3O^+]=1\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1%5Ctimes%2010%5E%7B-5%7DM)
Putting values in equation 1, we get:
![pH=-\log(1\times 10^{-5})](https://tex.z-dn.net/?f=pH%3D-%5Clog%281%5Ctimes%2010%5E%7B-5%7D%29)
![pH=5](https://tex.z-dn.net/?f=pH%3D5)
Now, putting values in equation 2, we get:
14 = 5 + pOH
pOH = 9
The solution is acidic in nature.