English
Karnataka Board PUCPUC Science Class 11

Assuming complete dissociation, calculate the pH of the following solution: 0.003 M HCl - Chemistry

Advertisements
Advertisements

Question

Assuming complete dissociation, calculate the pH of the following solution:

0.003 M HCl

Numerical
Advertisements

Solution

\[\ce{H_2O + HCl ↔  H_3O+ + Cl-}\] 

Since HCl is completely ionized,

`["H"_3"O"^+] = ["HCl"]`

`=>["H"_3"O"^+] = 0.003`

Now

`"pH" = - log["H"_3"O"^+] = - log(0.003)`

= 2.52

Hence, the pH of the solution is 2.52.

shaalaa.com
  Is there an error in this question or solution?
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×