Calculate the pH of a .700 M NaC2H3O2.
Ka (HC2H3O2)=1.8*10^-5
the answer is 9.29 but i dont know how it was solved by the packet
2007-01-28 11:33:07 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
NaC2H3O2 is the salt of a strong base (NaOH) with a weak acid (HC2H3O2). Thus the ion C2H3O2- will hydrolyze according to the reaction
.. .. .. .. .. .. C2H3O2- + H2O <=> HC2H3O2 + OH-
Initial .. .. .. .. 0.7
React .. .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. .. x
At Equil. .. .. 0.7-x .. .. .. .. .. .. .. .. .. .. x .. .. .. .. .. x
Kb = [HC2H3O2][OH-] / [C2H3O2-]
but also in such cases Kb=Kw/Ka (where Kw=10^-14 and it is the self dissociation constant of water), so you get
Kw/Ka= x^2/(0.7-x)
Le'ts assume that x << 0.7 and 0.7-x is practically equal to 0.7. Then the equation is simplified to Kw/Ka= x^2/0.7 =>
x= squareroot (0.7*Kw/Ka) = SQRT(0.7*(10^-14)/(1.8*10^-5)) =1.97*10^-5 which is truly <<0.7 so our assumption is correct (otherwise we would have to solve the quadratic equation)
pH=14-pOH= 14- (-logx)= 14-(-log(1.97*10^-5)) =9.29
2007-01-28 23:07:41 · answer #1 · answered by bellerophon 6 · 1⤊ 0⤋
Shucks, isn't it easier to use a pH meter or paper? That may be the theoretical answer but the proof is in the pudding.
2007-01-28 19:48:12 · answer #2 · answered by Anonymous · 0⤊ 4⤋
1 x 10^-14
------------- = 5.5 x 10^ -10
1.8 x 10 ^-5
-log 5.5 x 10^ -10 = 9.26
2007-01-28 19:45:19 · answer #3 · answered by davidosterberg1 6 · 2⤊ 2⤋
mmmmmm thats a difficult one to answer unless your smart good question though
2007-01-28 19:43:55 · answer #4 · answered by Anonymous · 0⤊ 4⤋