Is there any volume of 0.01 M CH3COOH given? I'll just assume a value for volume, so at least you can work out on your problem by following these steps.
Anyway, at equivalence point, all CH3COOH is converted to CH3COO-, thus the pH is determined by the concentration of CH3COO- ions.
The reaction involved is the hydrolysis of CH3COO- :
CH3COO- + H2O <===> CH3COOH + OH-
This reaction involves Kb, and we need to find the Kb. To solve, we use the formula
Kb = Kw / Ka
Substituting,
Kb = 1 x 10^-14 / 1.8 x 10^-5
Kb = 5.56 x 10^-10
I'll just assume a value for volume of 0.01 M CH3COOH to get the concentration of CH3COO-. For instance the the volume is 100 mL,
n,CH3COOH = 0.01 M 100 mL = 1 mmol CH3COOH
The moles of NaOH required:
1 mmol CH3COOH = 0.1 M * V,NaOH
V,NaOH = 10 mL NaOH
At equivalence point the total volume is
10 mL + 100 mL = 110 mL
and the concentration of CH3COO- ion is therefore,
1 mmol / 110 mL = 0.00909 M
Do the ICE table:
......CH3COO- | CH3COOH | OH-
I : 0.00909 | 0 | 0
C : -x | +x | +x
E : 0.00909-x | x | x
Set up the equation:
Kb = [CH3COOH][OH-] / [CH3COO-]
5.56 x 10^-10 = x^2 / (0.00909-x)
Assuming x << 0.00909, we can disregard it at the denominator:
5.56 x 10^-10 = x^2 / (0.00909)
Solving,
x = 2.24812 x 10^-6
pOH = -log(x)
pOH = -log(2.24812 x 10^-6)
pOH = 5.648
pH = 14 - pOH
pH = 8.35
Note this is not the answer (unless the given volume in the problem is the same as what I've assumed). You gotta work it out with the actual volume given
hope this helps~ `u`
calculate the ph at the equivalence point for the titration of 0.01 M Ch3COOH with 0.1 M NaoH ka= Ch3Cooh 1.8E-5
idek how to do it?
3 answers
jai i still need help
a solution containing 100.0 mL of 0.135 M CH3COOH (ka=1.8E-5) is being titrated with 0.54 M NaOH. calculate the ph:
at the equivalence point and 5 mL past the equivalence point.
how much is 5 mL past? i do not know how to do it?
at the equivalence point and 5 mL past the equivalence point.
how much is 5 mL past? i do not know how to do it?
0 answers