12.27

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RenuChepuru1L
Posts: 58
Joined: Thu Jul 27, 2017 3:00 am

12.27

Postby RenuChepuru1L » Fri Dec 08, 2017 7:47 pm

can someone explain part b to me?

Shannon Wasley 2J
Posts: 71
Joined: Fri Sep 29, 2017 7:06 am
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Re: 12.27

Postby Shannon Wasley 2J » 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


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