Asked by Anonymous
                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
                    Answered by
            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.
                    Answered by
            Anonymous
            
    why did you take 0.1M? Why don't you multify with volume in the first step?
    
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