What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water?
ACID =
with 8.00 mL of the 0.10 M HCl(aq)?
with 8.00 mL of the 0.10 M HC2H3O2(aq?
2 answers
im not sure Katie I'll tell you when i know but anyway its a great question i hope someone answers for you
1. HCl (strong acid completely split into ions)
*Find the concentration after dilution using:
V1M1 = V2M2
(8.00mL)(0.10 M) = (100.0 mL)(M2)
*M2 = (8.00mL)(0.10 M) / (100 mL) = ?______M
*pH = -log(M2)
2. HC2H3O2 (weak acid only partially dissociated into ions):
*Find the diluted concentration, M2 for HC2H3O2 as you did for HCl, then ..... let [H+] = x
x^2 / ( M2 - x ) = Ka (exact formula)
*x^2 / ( M2 ) = Ka (approximate formula, easier to use)
*Look up the Ka for acetic acid, HC2H3O2
*Substitute the values for M2 and Ka in to the approximate formula and solve for x
* pH = -log(x)
*Find the concentration after dilution using:
V1M1 = V2M2
(8.00mL)(0.10 M) = (100.0 mL)(M2)
*M2 = (8.00mL)(0.10 M) / (100 mL) = ?______M
*pH = -log(M2)
2. HC2H3O2 (weak acid only partially dissociated into ions):
*Find the diluted concentration, M2 for HC2H3O2 as you did for HCl, then ..... let [H+] = x
x^2 / ( M2 - x ) = Ka (exact formula)
*x^2 / ( M2 ) = Ka (approximate formula, easier to use)
*Look up the Ka for acetic acid, HC2H3O2
*Substitute the values for M2 and Ka in to the approximate formula and solve for x
* pH = -log(x)
1 answer