October 21, 2020, 12:20:16 PM
Forum Rules: Read This Before Posting


Topic: pH for titration-type question  (Read 3416 times)

0 Members and 1 Guest are viewing this topic.

Offline Zaphia

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
pH for titration-type question
« on: September 04, 2007, 01:05:22 AM »

I have been working on this problem for quite some time and was hoping that someone might have some suggestions on how to solve it:

"Calculate the pH of the final solution obtained after 200mL of 0.1 M NaOH is added to 300mL of 0.2 M H2SO4."
(H2SO4 + NaOH --> Na+ + HSO4- + HOH)

I started working the problem by obtaining the total moles of salt that was created (which came out to 0.06 moles). I then figured out the total moles of NaOH to be 0.02 moles (which is also the amount of salt created--limiting reactant).

I am not sure where to go from here, should I use the Henderson-Hasselhoff equation?? Or am I on the wrong train of thought because this reaction contains both a strong acid and base? The pK value for Sulfuric Acid is also confusing me, because it's a strong acid different references give me different values (-3.00, -4.80, etc).

~~I tried plugging in the numbers into the HH equation but I come up with a negative pH, which I am sure is incorrect.


Any suggestions would be much appreciated, thanks!

Online Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 26046
  • Mole Snacks: +1698/-402
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: pH for titration-type question
« Reply #1 on: September 04, 2007, 01:51:46 AM »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7553
  • Mole Snacks: +528/-88
  • Gender: Male
Re: pH for titration-type question
« Reply #2 on: September 04, 2007, 05:42:47 AM »
Calculate concentration of NaHSO4  and H2SO4 in the final solution.

As a first (and crude) approximation treat an excess of H2SO4 as a strong monoprotic acid.

As the better approximation solve a quadratic equation assuming an excessing H2SO4 dissociates completely on the first step of dissociation and partially on the second step (K2), even in the presence of HSO4- ions.
K2=(c+x)x/(c1-x)
where c is a concentration of the excess of H2SO4 after neutralisation,
c1 is a total molar concentration of H2SO4 in 500 mL solution (before neutralisation), and
x is the concentration of dissociated HSO4- and H3O+ from the  the second step dissociation of H2SO4 (K2)
Total concentration of H3O+ is c+x

Finally compare both approximations
(I guess the error will be about 0.2 pH unit).
AWK

Offline Zaphia

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Re: pH for titration-type question
« Reply #3 on: September 04, 2007, 10:03:54 AM »
Thank you for the input, very helpful :)

Sponsored Links