Question
sorry,for the mistake.
To 200mL of a 0.1M solution of a CH3COOH (Ka=1.8 x 10-5) 70mL of a 0.2M solution of NaOH have been added.calculate the pH before and after the NaOH addition.
To 200mL of a 0.1M solution of a CH3COOH (Ka=1.8 x 10-5) 70mL of a 0.2M solution of NaOH have been added.calculate the pH before and after the NaOH addition.
Answers
DrBob222
CH3COOH - acetic acid = HAc
CH3COONa = sodium acetate = NaAc
BEFORE: you have a 200 mL of 0.1M HAc. The pH is determined by the ionization of the HAc.
.........HAc ==> H^+ + Ac^-
I........0.1.....0......0
C.......-x.......x......x
E.......0.1-x....x......x
Substitute the E line into the Ka expression for HAc and solve for x = (H^+), then convert to pH.
AFTER.
millimols HAc = mL x M = 200 x 0.1 = 20
mmols NaOH added = 70 x 0.2 = 14
.....HAc + NaOH ==> NaAc + H2O
I....20.....0........0......0
add.........14...............
C...-14.....-14......14
E....6.......0.......14
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH. Post your work if you get stuck.
CH3COONa = sodium acetate = NaAc
BEFORE: you have a 200 mL of 0.1M HAc. The pH is determined by the ionization of the HAc.
.........HAc ==> H^+ + Ac^-
I........0.1.....0......0
C.......-x.......x......x
E.......0.1-x....x......x
Substitute the E line into the Ka expression for HAc and solve for x = (H^+), then convert to pH.
AFTER.
millimols HAc = mL x M = 200 x 0.1 = 20
mmols NaOH added = 70 x 0.2 = 14
.....HAc + NaOH ==> NaAc + H2O
I....20.....0........0......0
add.........14...............
C...-14.....-14......14
E....6.......0.......14
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH. Post your work if you get stuck.
Anonymous
why did you take 0.1M? Why don't you multify with volume in the first step?