Guest prvbspike Posted February 23, 2004 Share Posted February 23, 2004 Calculate the pH of 0.100 M sodium propanoate (NaC3H5O2)??? Link to comment Share on other sites More sharing options...
NSX Posted February 24, 2004 Share Posted February 24, 2004 hm...I'm not sure what NaC3H5O2 is, but when we had questions like these in CHem I, we know that it's a weak acid or base (since it's not a strong one). SO if it is a weak base or acid, the Na dissociates with the C3H5O2, and they undergo fwd. and bckwd. Rx until reaching an equilibrium. Usually, you need a Ka or Kb to solve these kinda' things, so you can figure out the concentration of H3O+. Then you can you the pH definition: pH = -log([H3O+]) Link to comment Share on other sites More sharing options...
wolfson Posted February 24, 2004 Share Posted February 24, 2004 Question: Calculate the pH of 0.100 M sodium propanoate (NaC3H5O2)? Answer: Ok since this is a base (weak) you need to use Kw: Kw = 1.0e-14 (e = x10 which is exp) Step 1: pH of base = Molarity / Kw pH of sodium propanoate = 0.100M / 1.0e-14 (Kw) = 1e13 (the new molarity) Step 2: (Log) pH = Log(new molarity) pH = Log(1e13) = 13 pH of 0.100M of Sodium Propanoate is 13. If you need anymore help don’t hesitate to ask. Link to comment Share on other sites More sharing options...
kitkitkit Posted February 24, 2004 Share Posted February 24, 2004 Cant you just use a pH testing method. I.e Universal Indicator Paper? Link to comment Share on other sites More sharing options...
YT2095 Posted February 24, 2004 Share Posted February 24, 2004 you can, but when asked to do it and show on paper (other than litmus LOL) you`de be in a mess after all who callibrated the indicators you use in the Lab? it all has to be mathematicly as well. I`m with you 100% in practical terms and I would stuggle to work it out on paper also, but it has to be done as part of lesson :? Link to comment Share on other sites More sharing options...
greg1917 Posted February 24, 2004 Share Posted February 24, 2004 pH calculations are a very routine part of chemistry and having to actually test everything for its pH value would be vastly inefficient and completely pointless. Link to comment Share on other sites More sharing options...
Gampin Posted February 24, 2004 Share Posted February 24, 2004 pH = 13 for a weak base? Link to comment Share on other sites More sharing options...
wolfson Posted February 24, 2004 Share Posted February 24, 2004 When saying "weak" base it does NOT mean the pH is weak (pH is only connected to the Molarity of a substance), a weak base is one which doesn't convert fully into hydroxide ions in solution, do not get confused try to remeber pH varies upon Molarity. Kb = [b:H+][OH-] / , If the question was relating to pKb the other considerations would need to come into affect. Hope you understand a little bit more now Link to comment Share on other sites More sharing options...
Hades Posted February 25, 2004 Share Posted February 25, 2004 im not disagreeing with u wolfson, but im rather confused. How can u calculate this problem without knowing what the Kb is? im assuming that our Kw = Kb to me. Also, considering that this problem reaches an equilibrium, i would determine whether my approximated value was less than 5%, if not, then i would have to solve using a quadratic equation based upon my unknowns (x) molarity. This is how i would have solved the problem, had i known the Kb: initial molarity of solution = .1 M My pretend rate : 1.8x10^-5 First, just as u did, Divide rate by molarity, and i come up with 1.8x10^-4. This number however is our unknown squared. I say this because we do not known what portion of this base is going to ionize, so we use variables. I cross multiplied : x^2 = 1.8x10^5 .100 - x 1 when we solve, we come up with the answer i gave, in its squared form. Its square route is .0013M or a pH of 7.36 b/c we approximated the value of x as a minute amount, we need to determine if it deviates more than 5% using this formula (H+)[u/] (initial Concentration of reactant) X 100 = 1.3, so it does not need to be solved via the quadratic equation. This is how i learned to calculate the ph of a weak base; if im wrong wolfson let me know, im not disputing, trying to learn that is all. Link to comment Share on other sites More sharing options...
Hades Posted February 25, 2004 Share Posted February 25, 2004 i can not edit my post, small correction x^2 = 1.8x10^5 .100 - x 1 should be x^2 = 1.8x10^5 .100------1.0 i dont know how the x got there Link to comment Share on other sites More sharing options...
wolfson Posted February 25, 2004 Share Posted February 25, 2004 Kw = Kb Not at all, Kw is the water constant, as i said in the previous post this is NOT calculating the pKb, we are calculating the pH. Kb = [b:H+][OH-] / Kw = 1.0e-14 (exp) Link to comment Share on other sites More sharing options...
Hades Posted February 26, 2004 Share Posted February 26, 2004 i understand that, yet i dont understand it clearly enough. i have my original chem notes from inorganic and the formula we used back then, the same as i used above was to calculate pH of a weak base that does not completely ionize. how can another formula be used? The equations u showed were what i would use to determine the pH of a strong acid or base that completely ionizes and does not leave behind any ambiguity of unknowns Link to comment Share on other sites More sharing options...
wolfson Posted February 26, 2004 Share Posted February 26, 2004 Ok we will calculate pKb (you are right that Kb is supposed to come intto affect FOR pKb usually in titrtion, so will will calculate pKb for sodium propanoate) Kb= 7.46e-10 (if i can remember correctly) xa = square root of [(7.46e-10) (0.100)] xa = 8.6e-6 M=[OH] xa / Kw = Mb Mb = 8.6e-6 / 1.0e-14 = 8.6e8 Log(Mb) = pKb-pOH pKb = Log(806e8) = 8.93 pKb = 8.93. So we can see that using pKb we therefore have pKb calculated a weak base, is this what you mean, i would actually go with pKb. Link to comment Share on other sites More sharing options...
wolfson Posted February 26, 2004 Share Posted February 26, 2004 Hades, oh my, im such a fool, the pKb is the pH of a weak base, im so sorry Hades, and i have a BSc in chemistry how, lol i do appologise use the above it is the correct way, i was still thinking strong base lol. Link to comment Share on other sites More sharing options...
agaubr Posted February 27, 2004 Share Posted February 27, 2004 PKb is the log [ 1/Kb] where is the equilibrium constant for a base which tells the extend of inioization of the base and the increase in the [OH-] or decrease in the [H+]. A week base would have a pH greater than 7 but less than 9. A pH of 13 is a very strong base since [H+] is very low and mostly [OH-] present in solution. Ag Link to comment Share on other sites More sharing options...
wolfson Posted February 27, 2004 Share Posted February 27, 2004 Thus: Kb = antilog ( - pKb) Link to comment Share on other sites More sharing options...
wizzer1223 Posted April 18, 2007 Share Posted April 18, 2007 Im lost about what wolfson did. when he got into what the actual Kb value was I could solve it but his Molarity/Kw didnt work at all. Link to comment Share on other sites More sharing options...
JayLechner Posted June 24, 2008 Share Posted June 24, 2008 I dont know if this helps at all but pH stands for percent of hydronium ions. Link to comment Share on other sites More sharing options...
hermanntrude Posted June 25, 2008 Share Posted June 25, 2008 I dont know if this helps at all but pH stands for percent of hydronium ions. no it doesnt. the small p in pH represents the negative log base ten. pH = -log [H30+] Link to comment Share on other sites More sharing options...
Ladeira Posted June 28, 2008 Share Posted June 28, 2008 (edited) Cant you just use a pH testing method. I.e Universal Indicator Paper? It depends on the situation. I also use the Universal Indicator Paper because I've just ended General Chem II in Chemistry Technician school. But for researchers and college when they want deep results, they must use that famous math: logarythms. I don't know how to use logarythms for pH (I'll learn it soon), but I think it's more precise and more attractive, if you figure out what I mean. Edited June 28, 2008 by Ladeira I didn't make myself clear. Link to comment Share on other sites More sharing options...
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