HomeworkQuestion 13.11 part b

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HomeworkQuestion 13.11 part b

Postby 904475052_ » Fri Dec 04, 2015 10:37 am

When you add 20.0ml (.020L) of .100M HNO3 to the initial buffer (acetic acid with .100M and 100.ml or .100L) what is the ph change? So first you find the concentration of HNO3 and then than of acetic acid but I got confused in the solution how come they add 1.00x1^-2 to the new mol of acetic acid. Where does it come from?

Leena Tran 2K
Posts: 20
Joined: Fri Sep 25, 2015 3:00 am

Re: HomeworkQuestion 13.11 part b

Postby Leena Tran 2K » Fri Dec 04, 2015 7:25 pm

The 1.00 x 10^-2 mol of acetic acid is from the initial concentration of acetic acid, 0.100 M acetic acid in a solution with a volume of 100 mL. It is added to the 2.00 x 10^-3 mol of acetic acid made with the addition of the HNO3. From there, you can calculate the concentrations of acetic acid and the acetate ion, and plug them into the Henderson-Hasselbalch equation to find the pH.

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