### Question 12.19 (6th Edition)

Posted:

**Sat Dec 01, 2018 5:06 pm**The concentration of HCl is reduced to 12% of its initial value by dilution. By what amount does the pH of the solution change.

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=60&t=37982

Page **1** of **1**

Posted: **Sat Dec 01, 2018 5:06 pm**

The concentration of HCl is reduced to 12% of its initial value by dilution. By what amount does the pH of the solution change.

Posted: **Sat Dec 01, 2018 6:40 pm**

Original concentration: 10^{x}

Concentration diluted to 12% of original value: (10^{x})(0.12)

Original pH: -log(10^{x} )

pH after dilution: -log((10^{x})(0.12)) = -log(10^{x}) - log(0.12)

Change in pH = (Original pH) - (pH after dilution) = -log(10^{x}) - (-log(10^{x}) - log(0.12))

Change in pH = -log(10^{x}) + log(10^{x}) - log (0.12)

Change in pH = -log(10^{x}) + log(10^{x}) - log (0.12)

Change in pH = - log (0.12)

Change in pH = 0.92

As a shortcut, you could just remember that the change in pH is the negative log of the dilution percentage.

Concentration diluted to 12% of original value: (10

Original pH: -log(10

pH after dilution: -log((10

Change in pH = (Original pH) - (pH after dilution) = -log(10

Change in pH = -log(10

Change in pH = -log(10

Change in pH = - log (0.12)

Change in pH = 0.92

As a shortcut, you could just remember that the change in pH is the negative log of the dilution percentage.