Chemical Forums

Please login or register.

Login with username, password and session length

Sponsored links

Pages: [1]   Go Down

Author Topic: Chemistry Equilibrium Question Help  (Read 675 times)

0 Members and 1 Guest are viewing this topic.

kraver0

  • New Member
  • **
  • Mole Snacks: +0/-0
  • Offline Offline
  • Posts: 3
Chemistry Equilibrium Question Help
« on: November 29, 2012, 04:16:02 PM »

A solution that contains 1.48 M hydrofluoric acid, HF (Ka = 7.2E-4), and 3.00 M hydrocyanic acid, HCN (Ka = 6.2E-10).

What is the pH of this mixture of weak acids?

a) 1.49
b) 2.97
c) 4.52
d) 9.04
e)12.51

The correct answer is a. But I don't know how to get this answer. Can someone help me? Thanks!
Logged

Hunter2

  • Sr. Member
  • *****
  • Mole Snacks: +98/-33
  • Online Online
  • Gender: Male
  • Posts: 1264
  • Vena Lausa moris pax drux bis totis
Re: Chemistry Equilibrium Question Help
« Reply #1 on: November 29, 2012, 09:17:44 PM »

Check this:

its in german but the formulas you can read

http://www.chemieunterricht.de/dc2/mwg/ph-saeuremischung.htm
Logged

AWK

  • Global Moderator
  • Sr. Member
  • ***
  • Mole Snacks: +358/-108
  • Offline Offline
  • Gender: Male
  • Posts: 4861
Re: Chemistry Equilibrium Question Help
« Reply #2 on: November 29, 2012, 09:57:34 PM »

Could you calculate pH based on the HF present in this solution? This is quite sufficient result in this case.
Logged
AWK

vin

  • New Member
  • **
  • Mole Snacks: +0/-0
  • Offline Offline
  • Posts: 6
Re: Chemistry Equilibrium Question Help
« Reply #3 on: December 02, 2012, 05:17:15 AM »

use the formula [H+] = square root (Ka x C)
to find out the concentrations of H+ of both the acids. Total the concentrations and use the formula pH = -log[H+]
You get the answer....
step - 1
H+ ions from HCN = 4.31 x 105-
step - 2
H+ ions from HF = 3268.66 x 105-
Add the above concentrations and find negative log... you get 1.48
Hope this helped you.... good luck
Logged

Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Mole Snacks: +1387/-370
  • Offline Offline
  • Gender: Male
  • Posts: 21266
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Chemistry Equilibrium Question Help
« Reply #4 on: December 02, 2012, 07:01:26 AM »

use the formula [H+] = square root (Ka x C)
to find out the concentrations of H+ of both the acids. Total the concentrations and use the formula pH = -log[H+]
You get the answer....
step - 1
H+ ions from HCN = 4.31 x 105-
step - 2
H+ ions from HF = 3268.66 x 105-
Add the above concentrations and find negative log... you get 1.48
Hope this helped you.... good luck

And it is completely wrong.
Logged
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

vin

  • New Member
  • **
  • Mole Snacks: +0/-0
  • Offline Offline
  • Posts: 6
Re: Chemistry Equilibrium Question Help
« Reply #5 on: December 02, 2012, 07:24:41 AM »

can you tell how it is wrong?
Logged

Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Mole Snacks: +1387/-370
  • Offline Offline
  • Gender: Male
  • Posts: 21266
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Chemistry Equilibrium Question Help
« Reply #6 on: December 02, 2012, 07:31:29 AM »

First of all - you can't simply sum concentrations, as H+ produced in one reaction shifts the equilibrium in the other reaction.

Second - you are using an approximate method not checking if it can be used.
Logged
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info
Pages: [1]   Go Up
 

Mitch Andre Garcia's Chemical Forums 2003-Present.

Page created in 0.196 seconds with 23 queries.