Acetate Approx
is this pH calculation correct (buffers)?
a solution 50mL of a .1M sodium acetate solution with 150mL of 1M acetic acid(pKa = 4.7) what is the pH of the resulting solution.
first i calculated molality of the solution[ .1/.05L] = 2M and [1/.15] = 6.7M
next to convert pKa to Ka, 10^-4.7= 2.0 x 10^-5
now to set it up.(i did an ice table)
2.0 x 10^-5 = [(x)(6.7 +x)]/2-x), for this step (6.7+x) is aprrox 6.7 and (2-x) is approx 2
2.0 x 10^-5 = x6.7/2 = 3.35x
x = 6×10^-6
then -log(6×10^-6) = 5.2 pH
is it right ?
Nope, it’s not correct.
Molarity calculations (use M1V1 = M2V2):
NaAc: (0.1) (50) = (x) (200); x = 0.025 M
HAc: (1) (150) = (x) (200); x = 0.75 M
Use Henderson-Hasselbalch equation to calculate pH of buffer:
pH = pKa + log (salt/acid)
x = 4.75 + log(.025/.75)
x = 4.75 -1.48
pH = 3.27
http://www.chemteam.info/AcidBase/AcidBase.html
Update: note that I used a pKa of 4.75, not the 4.7 you supplied.
Precipitation of Na-acetate from a supersaturated solution
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