[777888]

I don't know if I did it right! It is quite important for me! Can somebody see if my method and assumption are OK? THANKS

1) 50mL of 0.1 M CH3COOH (Ka=1.8x10^-5) is titrated by 0.3 M NaOH. Calculate the pH of the solution after 18mL 0.3 M NaOH has been added

MY CALCULATIONS:
CH3COOH + NaOH -> NaCH3COO + H2O

CH3COOH:
n=cv=5x10^-3mol
V=0.050L

NaOH:
V=0.018L
c=0.3mol/L
n=cv=5.4x10^-3mol

mol NaOH remaining=5.4x10^-3 - 5x10^-3

[OH-]=n/V(total)=4x10^-4 / 0.068 = 5.8824x10^-3 M
pOH=2.2304
pH=11.7696=11.8

NOTE:
For the equilibrium:
CH3COO- + H2O <-> CH3COOH + OH-
Since the Kb value of this equilibrium is so low (Kb of CH3COO- = 5.6x10^-10), it contibutes an insignificant amount of OH- ions to the solution. Also the relatively large amounts of OH- entering the solution shift the equilibrium to the left, reducing even farther the tiny amount of OH- it produces.

I have assumed that the amount of OH- produced by the equilibrium reaction is negligible. AM I RIGHT? Is this assumption OK?

If not, how can I calculate the required pH?
THANK YOU

[AWK]

1) 0.1 M CH2COOH (Ka=1.8x10^-5) is titrated by 0.3 M NaOH. Calculate the pH of the solution after 18mL 0.3 M NaOH has been added

should be CH3COOH, its volume not stated at the beginning

n=c/v=5x10^-3mol

n = c x V (multiplying not dividing) - though this result and the rest of calculations are correct

[777888]

I mistyped them...

So is my assumption OK? (I have assumed that the amount of OH- produced by the equilibrium reaction is negligible)

Thanks!