find the ph of 0.100M NaC2H3O2 solution.
do i just write an equation
and solve for x
NaC2H3O2 +H2O= Na2O + HC2H3O2-
0.100 0 0
so x^2/0.100-x =1.8 X10^-5
x=1.3X10^-3
then do ph=-log(1.3X10-3)
ph=2.9
is this right?
2 answers
probably right
I know acetate solutions have a pH of about 8 or so; therefore, this must not be right. What you wrote is close but then you must hydrolyze the acetate ion. If we call acetate, Ac^-, then
..........Ac^- + HOH ==> HAc + OH^-
initial....0.1............0......0
change......-x............x.......x
equil....0.1-x............x........x
Kb for Ac^- = (Kw/Ka for HAc) = (HAc)(OH^-)/(Ac^-)
Substitute and solve for x = OH^-, then convert to pH. Your answer is about the pH of 0.1M acetic acid. This should come out around 8.
..........Ac^- + HOH ==> HAc + OH^-
initial....0.1............0......0
change......-x............x.......x
equil....0.1-x............x........x
Kb for Ac^- = (Kw/Ka for HAc) = (HAc)(OH^-)/(Ac^-)
Substitute and solve for x = OH^-, then convert to pH. Your answer is about the pH of 0.1M acetic acid. This should come out around 8.