Calculating titrations beyond stoichiometric point

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Ashley Chen
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

Calculating titrations beyond stoichiometric point

Postby Ashley Chen » Wed Dec 02, 2015 9:22 pm

In homework, 13.27 part e, it asks calculate the pH at each stage in the titration for the addition of 0.150M HCL to 25.0mL of 0.110M NaOH after the addition of 5.0mL of acid beyond stoichiometric point.

Why does the calculation only include [H30+]? I'm confused on the reasons to why the calculations are the way they are in the solution manual. can someone explain?

Brandon Truong 2G
Posts: 29
Joined: Fri Sep 25, 2015 3:00 am

Re: Calculating titrations beyond stoichiometric point

Postby Brandon Truong 2G » Wed Dec 02, 2015 11:05 pm

Because at the stoichiometric point, all of the base has been neutralized by the acid and leaving behind only a salt and water. Since NaOH is a strong base and HCl are strong acids, the salts that are formed do not affect the pH. When you keep adding more acid, all you're doing is adding more H30+ since there is no more OH- to neutralize it.

Alison Ou 14a
Posts: 32
Joined: Fri Sep 25, 2015 3:00 am

Re: Calculating titrations beyond stoichiometric point

Postby Alison Ou 14a » Wed Dec 02, 2015 11:26 pm

My reply is attached as an image since it's easier to write equations down than to type them. Hope it helps!
Attachments
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Pauline Tze 3B
Posts: 57
Joined: Sat Jul 09, 2016 3:00 am
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Re: Calculating titrations beyond stoichiometric point

Postby Pauline Tze 3B » Fri Dec 02, 2016 1:21 am

This was really helpful! Thank you Alison :)


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