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Can anyone who is good at chem ap please help me do #1 on this site:

http://www.collegeboard.com/prod_downloads/ap/students/chemistry/chemistry_frq_02.pdf

The question is on page 6. It has 5 parts. I was studying from it and wondering whether or not I got them right. Thank you so much.

By Jason817 (Jason817) on Saturday, March 01, 2003 - 04:28 pm: Edit

anyone?

By Yourlocalmayor (Yourlocalmayor) on Saturday, March 01, 2003 - 06:10 pm: Edit

M1V1=M2V2, eh, how about you post your answers and then we'll see if you got em right?

By Absurdistx (Absurdistx) on Saturday, March 01, 2003 - 06:26 pm: Edit

I think we've done this problem before in AP Chem. Not sure of all these answers, but you can decide if they seem right.

A. pH = 4.95, -log(4.95) = 1.12E-5. So [H+] = 1.12E-5 M

B. Ka = [H+][OBr-] / [HOBr]
2.3E-9 = (1.8E-5)² / (x-1.8E-5)
x = 1.4E-1 M

C. i.)
(.065)(.146) = 9.49E-3 moles HOBr
(x)(.115) = 9.49E-3 moles Ba(OH)2; x = 8.25E-2 liters
ii.) Greater. Weak acid, strong base, so higher pH.

D. pH = pK + log [A-/HA]
8.30 = 8.63 + log [A-/.160]
[A-] = 7.34E-2 M * (.125 L) = 9.18E-3 moles
*Not really sure of this one, perhaps you need an ICE chart?

E. More oxygens atoms mean a strong acid. Oxygen atoms are extremely electronegative. The greater number of oxygens in the molecule, the greater the shift of electrons toward the oxygens and the weaker the bond between the hydrogens bonded to oxygens becomes.

By Jason817 (Jason817) on Saturday, March 01, 2003 - 06:39 pm: Edit

thank you. I dont have my answers with me though.

Absurdistx:

For part A, when you are given the pH, don't you use anti-log, 10^(-pH), to find the concentration. If the concentration is given, then you take -log(concentration) to get pH.

By Jason817 (Jason817) on Saturday, March 01, 2003 - 06:42 pm: Edit

Also, can you explain the second part of Part C. I don't get that one.

By Absurdistx (Absurdistx) on Saturday, March 01, 2003 - 07:13 pm: Edit

You're right about the 10^(-pH)...I did that and put the answer, but for some reason, I put -log(pH).
And for part C, isn't the equivalence point where moles acid = moles base added? So I found moles of acid and set up an equation to find the volume of base, since we knew we'd need 9.49E-3 moles of it.

By Eurostar (Eurostar) on Friday, March 21, 2003 - 04:10 am: Edit

http://apcentral.collegeboard.com/repository/sg_chemistry_02_11388.pdf

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