posted by rick .
propanoic acid ch3ch2cooh is a weak acid which has a value for dissociation content of ka = 6.3 x 10^-6 calculate the pH of an aqueous solution of propanoic acid containing 37 g of propanoic acid per litre.
i'm stumbled on this one.. i calculated that the concentration of the propanoic acid is .635 mol/L but i don't know what to do after.. do i just do -log(.635)?? thanks for any help!
someone? anyone? help!
Not quite. Let's call propionic acid (Propanoic acid) HPr to avoid all that typing.
HPr ==> H^+ + Pr^-
K = (H^+)(Pr^-)/(HPr)
Now, we have 37 g HPr/L which is 37/74 = 0.50 mole/L = 0.50 M.
Set up an ICE chart.
(H^+) = x
(Pr^-) = x
(HPr) = 0.5-x
Substitute into Ka expression and solv for (H^+). My answer is 0.00177 M. Then pH = -log(1.77 x 10^-3) and round to 3 s.f. for about 2.75 but you need to confirm all of that.
thank you very much. i did 47/74 instead of 37/74. I knew the steps to answer the question but the confusing part was forming the balanced equation because i didn't know how to do it and the question didn't offer one. May I ask how you got that? how do you know that H+ when balanced has the same amount of moles as ch3ch2cooh?
First, you need to remember that ch2ch2cooh doesn't mean a thing to a chemist.
CH3CH2COOH is the formula for propanoic acid. It is the last H on the right hand side of the formula that is the acidic hydrogen. All of the others are glued onto their respective C atoms and they aren't going anywhere. In organic chemistry, the -COOH group is acidic because the H can ionize. Most of the COOH compounds are relatively weak acids. So we write RCOOH to show any organic acid. I just used the right hand H as H and stuck Pr on for the R(the rest of the molecule) because it is so much easier to write HPr than
CH3CH2COOH ==> CH3CH2COO^- + H^+.
Thank you sooo much!