HomeworkQuestion 13.11 part b

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904475052_
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HomeworkQuestion 13.11 part b

Postby 904475052_ » Fri Dec 04, 2015 10:36 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?

Andrea Sandri 3D
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

Re: HomeworkQuestion 13.11 part b

Postby Andrea Sandri 3D » Fri Dec 04, 2015 11:06 am

They added 0.01 to the new mol of acetic acid because it is specified in the text of the question that "A buffer solution of volume 100.0 mL is 0.100 M CH3COOH(aq) and 0.100 M NaCH3CO2(aq)." This means that to find the number of moles of acetic acid and acetate, you would have to multiple their respective concentrations (0.100 M and 0.100 M) by the volume (0.100 L), which results in 0.01 moles. Then since you know that HNO3 is a strong acid and thereby produces CH3COOH from CH3CO2-, you must add the moles of acetic acid produced by the addition of HNO3 (0.002) to the initial number of moles of acetic acid (0.01). Similarly, you must subtract the moles of acetic acid produced by the addition of HNO3 (0.002) from the initial moles of acetate (0.01).


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