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?
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`
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?
To calculate the pH at the equivalence point of a titration, we need to understand the concept of equivalence point and how it relates to the reaction between a weak acid (CH3COOH) and a strong base (NaOH).
At the equivalence point, the number of moles of acid is equal to the number of moles of base. This means that all the acetic acid (CH3COOH) will react with the sodium hydroxide (NaOH) in a 1:1 ratio.
Here's how we can calculate the pH at the equivalence point:
1. Write the balanced chemical equation for the reaction between CH3COOH and NaOH:
CH3COOH + NaOH -> CH3COONa + H2O
2. Calculate the number of moles of acetic acid (CH3COOH) in the initial solution:
Moles of CH3COOH = Molarity (0.01 M) × Volume (given or measured in liters)
3. Since the reaction is 1:1, the number of moles of NaOH needed to neutralize the acetic acid is equal to the number of moles of CH3COOH.
4. Calculate the volume of NaOH required using the stoichiometry of the reaction:
Volume of NaOH (in liters) = Moles of CH3COOH × Volume factor of NaOH/1 mole of CH3COOH
5. Now, calculate the concentration of CH3COONa formed at the equivalence point:
Moles of CH3COONa = Moles of CH3COOH
Concentration of CH3COONa at equivalence point = Moles of CH3COONa / Total volume of the solution
6. CH3COONa is a salt of a weak acid (CH3COOH), so we need to consider its hydrolysis reaction:
CH3COO- + H2O ⇌ CH3COOH + OH-
At the equivalence point, CH3COONa will fully hydrolyze into CH3COOH and OH- ions.
7. The OH- ions will react with water to form hydroxide ions (OH-) and increase the pH. To calculate the concentration of OH- ions, we need to use the reaction quotient (Q), which is the initial concentration of CH3COO- ions divided by the initial concentration of acetic acid (CH3COOH) before the reaction.
8. Calculate the concentration of OH- ions using the reaction quotient:
Q = [CH3COOH] / [CH3COO-]
Concentration of OH- ions = Ka × Q / (1 + Ka)
9. Finally, calculate the pOH at the equivalence point:
pOH = -log10[OH- concentration]
10. Calculate the pH at the equivalence point:
pH = 14 - pOH
By following these steps and using the given values, you should be able to calculate the pH at the equivalence point for the titration of CH3COOH with NaOH.