### 12.27

Posted:

**Fri Dec 08, 2017 7:47 pm**can someone explain part b to me?

Created by Dr. Laurence Lavelle

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

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=57&t=24845

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Posted: **Fri Dec 08, 2017 7:47 pm**

can someone explain part b to me?

Posted: **Fri Dec 08, 2017 10:41 pm**

For part b, the first step would be to use the formula:

M(initial)V(initial) = M(final)V(Final)

The initial M is given: .025

The initial volume is given: 200mL --> .200L

The final volume is also given: 250mL --> .250

Therefore by solving (.200)(.025) = (.250)x, you get the final concentration is equal to .02M HCl. However, since HCl is a strong acid, it dissociates completely, so the concentration of Hal is equal to the concentration of H3O+. So you can use the value of [H3O+] = .02 and solve -log(.02) to get the pH of 1.7

M(initial)V(initial) = M(final)V(Final)

The initial M is given: .025

The initial volume is given: 200mL --> .200L

The final volume is also given: 250mL --> .250

Therefore by solving (.200)(.025) = (.250)x, you get the final concentration is equal to .02M HCl. However, since HCl is a strong acid, it dissociates completely, so the concentration of Hal is equal to the concentration of H3O+. So you can use the value of [H3O+] = .02 and solve -log(.02) to get the pH of 1.7