0.400 g/molar mass propionic acid gives you the moles of propionic acid that you started with. Now that is titrated with 0.150 M NaOH. How much NaOH will it take. Moles NaOH will be same as the moles of propionic acid (since they react in 1:1 ratio) so moles NaOH you know and that = M x L. Knowing moles and M you calculate L NaOH. So the salt produced from the titration: Letting HP stand for propionic acid:
HP + NaOH ==> NaP + HOH.
You can do the ice table if you wish to determine how much NaP is present at the equivalence point but I can tell you before you do it that moles NaP will be the same as the moles HP you started with. So what is the concentration of the NaP? That will be moles NaP/volume at that point (take the volume of NaOH and add to 50 mL--the volume initially used.) The pH at the equivalence point is determined by the hydrolysis of the salt (in this case the NaP). So write a hydrolysis equation, set up an ICE table, and solve for OH^-.
P^- + HOH ==> HP + OH^-
ICE table here.
Kb for P^- is Kw/Ka for propioinic acid.
Question: A 0.400 g sample of propionic acid was dissolved in water to give 50.00 mL of solution. This solution was titrated with 0.150 M NaOH. what was the pH of the solution when the equivalence point was reached?
I'm not really sure how i would go about setting this up. Would i use an ice table? What do i do with the 0.400 g
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