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Ph question ( a little more advanced)?
What is the Ph of the solution which results from mixing 150 ml of 0.50 M Ch3CO2H(aq) and 150 ml of 0.50 M NaOH(aq) at 25 degree's Celsius? (Ka of CH3CO2H=1.8 X 10^-5)
I know this may be tedious for you to do, but if you can put your work, so i can see what you did to achieve your answer. Thanks so much to who ever will do this. Thanks sooo much.
1 Answer
- Dr.ALv 71 decade agoFavorite Answer
moles acetic acid = 0.150 L x 0.50 M=0.075
moles NaOH added = 0.150 x 0.50 M = 0.075
CH3COOH + NaOH = CH3COONa + H2O
moles sodium acetate formed = 0.075
total volume = 0.300 L
[acetate]= 0.075 / 0.300 L=0.25 M
sodium acetate is a strong salt : CH3COONa = CH3COO- + Na+
CH3COO- + H2O < => CH3COOH + OH-
Kb = Kw/Ka = 5.6 x 10^-10 = x^2 / 0.25-x
x = [OH-]= 1.18 x 10^-5 M
pOH = 4.93
pH = 9.07
