Chapter 13 #35

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Michael Lonsway 3O
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Joined: Wed Sep 21, 2016 2:57 pm

Chapter 13 #35

Postby Michael Lonsway 3O » Sun Nov 27, 2016 5:18 pm

Suppose that 25.0ml of .10M CH3COOH is titrated with .10M NAOH. a)What is the initial PH of the .10M CH3COOH solution? b) What is the PH after the addition of 10.0ml of .10M NAOH?c)What volume of .10M NaOH is required to reach halfway to the stoichiometric point? d) Calculate the PH at that halfway point. e)What volume of 0.10M NaOH is required to reach the stoichiometric point? f) Calculate the PH at the stoichiometric point.

I'd like help with parts c and e since I'm not sure how to find volume needed when given the concentration of the acid and base.

Liam Giffin 2B
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Re: Chapter 13 #35

Postby Liam Giffin 2B » Wed Nov 30, 2016 10:07 pm

I also had trouble with this problem. In part B the solutions manual finds the concentrations of CH3COOH and CH3CO2- after neutralization, which makes sense to me. Then it sets up an ICE box to solve for the hydronium ion concentration. This works, but I also got the same answer when I plugged in the values for the concentrations into the Henderson-Hasselbach equation. Do we need to set up the ice box or can we just use the Henderson-Hasselbach equation?

Kylee 3f
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Joined: Wed Sep 21, 2016 2:57 pm

Re: Chapter 13 #35

Postby Kylee 3f » Sat Dec 03, 2016 6:32 am

How did you find the (1.5x10^-3)? the after neutralization part?

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Joined: Fri Jul 15, 2016 3:00 am

Re: Chapter 13 #35

Postby Amy_Shao_2D » Sat Dec 03, 2016 1:28 pm

I think the 1.5x10^-3 is from 2.5x10^-3 mol CH3COOH reacting with 1.0x10^-3 mol OH-, so you subtract 1.0x10^-3 from 2.5x10^-3

But then I don't get how 1.0x10^-3 mol OH- is changed to 1.0x10^-3 mol CH3CO2...

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