November 27, 2024, 01:53:35 AM
Forum Rules: Read This Before Posting


Topic: Questions about HOCL as an acid  (Read 7936 times)

0 Members and 2 Guests are viewing this topic.

imcuteali2

  • Guest
Questions about HOCL as an acid
« on: February 19, 2006, 08:12:27 PM »
(1) Calculate the pH of a 0.10 M solution of HOCl, Ka = 3.5 x 10^-8.
A. 4.23
B. 8.46
C. 3.73
D. 1
E. 3.23

Answer A, 4.23 because Ka=3.5 x 10^-8= [H][OCL-]/ [HOCL]
Initial HOCL=.1M
use ice table procudure to obtain:
Ka=3.5x10^-8= (x)^2.(.10-x)
Since K value is so small we can ignore x term in the denominator.
x= 5.9x10^-5

check if error is under 5%
(5.9x10^-5/.10 )x 100%= .059%
therefore pH= -log(concentration)= 4.23

(2) The following question refers to a 0.70 M solution of hypochlorous acid, HClO.
If the molarity was decreased to 0.3 M, which of the following statements would be true?

A. The % dissociation would not change.
B. The % dissociation would increase.
C. The % dissociation would decrease.
D. The equilibrium constant would stay the same.
E. Two of these.

I think it is E because even though the concentration decreased, the overall percentage of dissociation would stay the same. Also, the question does not say that the temperature is changing therefore the answer is also D...therefore 2 of these are correct....Am I correct on this?

Thank you for you *delete me*

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27862
  • Mole Snacks: +1813/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re:Questions about HOCL as an acid
« Reply #1 on: February 20, 2006, 02:49:01 AM »
pH= -log(concentration)= 4.23

OK

Quote
I think it is E because even though the concentration decreased, the overall percentage of dissociation would stay the same. Also, the question does not say that the temperature is changing therefore the answer is also D...therefore 2 of these are correct....Am I correct on this?

It is E, but you are wrong. Dissociation constant remains the same, but dissociation % will change, thus one of B/C will be correct. More precisely - diluting solution you are increasing dissociation % - infinitesimally diluted solution is ionized 100%.
« Last Edit: February 20, 2006, 03:01:31 AM by Borek »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links