1.
HC4H7O2 + H2O ==> H3O^+ + C4H7O2^-
2.
pH = -log(H3O^+)
-2.56 = log(H3O^+)
(H3O^+) = about 2.8E-3 but you need to do it more accurately. Let HB stand for butyric acid, then
.......HB + H2O ==> H3O^+ B^-
I.....0.5M...........0.....0
C......-x...........x.......x
E.....0.5-x.........x.......x
Ka = (H3O^+)(B^-)/(HB)
You know (H3O^+) = about 2.8E-3, that's also B^-. Substitute into Ka expression and solve for Ka.
3.
C4H7O2^- + H2O ==> H3O^+ + HC4H7O2
4.
Kb for B^- = (Kw/Ka for HB)
5. Use the Henderson-Hasselbalch equation.
Butyric acid (HC4H7O2) is a weak acid with the stench of rancid butter
1.) write the equation for the dissociation of HC4H7O2 in water
2.)the pH of .5 M HC4H7O2 is 2.56 calculate Ka of butyric acid
3.) Write the equation for the hydrolysis of C4H7O2-
4.) Calculate Kb for C4H7O2-
5.) Calculate the pH of a buffer solution in which the concentration of butyri acid= .36 M and the concentration of C4H7O2- is .31 M.
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