Please help me with the following task:
20.0 mL of acetic acid solution is titrated with 0.0510 M sodium hydroxide solution.
a) Calculate the concentration of acetic acid if the pH at the equivalence point is 8.51! Assume additivity volume.
Ka(CH3COOH) = 1.75·10-5
b) Calculate the volume of NaOH added.
Many thanks in advance!
Titration
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- ChenBeier
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Re: Titration
pH = ½ (pka + log cHAc + 14)
Solve for cHAc
Solve for cHAc
Re: Titration
To calculate the concentration of acetic acid and the volume of NaOH added, we can use the information given and perform the following steps:
a) Calculate the concentration of acetic acid (CH3COOH):
At the equivalence point of the titration, the moles of acetic acid (CH3COOH) will be equal to the moles of sodium hydroxide (NaOH) added.
The volume of NaOH added at the equivalence point can be calculated using the formula: Volume (NaOH) = Concentration (NaOH) * Volume (NaOH solution).
The moles of NaOH can then be determined using the equation: Moles (NaOH) = Concentration (NaOH) * Volume (NaOH).
Since the moles of NaOH and CH3COOH are equal at the equivalence point, we can set up the following equation:
Moles (CH3COOH) = Moles (NaOH)
Concentration (CH3COOH) * Volume (CH3COOH solution) = Concentration (NaOH) * Volume (NaOH)
Now, we can plug in the given values:
Concentration (NaOH) = 0.0510 M (given)
Volume (NaOH solution) = Volume of NaOH added = to be determined
Concentration (CH3COOH) = to be determined (we need to find this)
Let's calculate the concentration of acetic acid (CH3COOH):
Using the equation above:
Concentration (CH3COOH) = (Concentration (NaOH) * Volume (NaOH solution)) / Volume (CH3COOH solution)
We can rearrange the equation to solve for the concentration of acetic acid (CH3COOH):
Concentration (CH3COOH) = (Concentration (NaOH) * Volume (NaOH solution)) / Volume (CH3COOH solution)
Given pH at the equivalence point is 8.51, we know that pOH = 14 - pH = 14 - 8.51 = 5.49.
From this pOH value, we can determine the hydroxide ion concentration ([OH-]) at the equivalence point.
[OH-] = 10^(-pOH) = 10^(-5.49)
Next, we can use the equilibrium constant expression for acetic acid to find its concentration:
Ka = [H3O+][CH3COO-] / [CH3COOH] mymilestonecard
[H3O+] = [OH-] (at equivalence point, the solution is neutral)
Now, we can solve for the concentration of acetic acid (CH3COOH):
[CH3COOH] = [OH-] = 10^(-5.49)
b) Calculate the volume of NaOH added:
Now that we have found the concentration of acetic acid (CH3COOH), we can use the equation:
Volume (NaOH solution) = (Concentration (CH3COOH) * Volume (CH3COOH solution)) / Concentration (NaOH)
Plug in the values:
Volume (NaOH solution) = (Concentration (CH3COOH) * 20.0 mL) / 0.0510 M
Now, we can substitute the value of [CH3COOH] that we found in step (a) to calculate the volume of NaOH added.
a) Calculate the concentration of acetic acid (CH3COOH):
At the equivalence point of the titration, the moles of acetic acid (CH3COOH) will be equal to the moles of sodium hydroxide (NaOH) added.
The volume of NaOH added at the equivalence point can be calculated using the formula: Volume (NaOH) = Concentration (NaOH) * Volume (NaOH solution).
The moles of NaOH can then be determined using the equation: Moles (NaOH) = Concentration (NaOH) * Volume (NaOH).
Since the moles of NaOH and CH3COOH are equal at the equivalence point, we can set up the following equation:
Moles (CH3COOH) = Moles (NaOH)
Concentration (CH3COOH) * Volume (CH3COOH solution) = Concentration (NaOH) * Volume (NaOH)
Now, we can plug in the given values:
Concentration (NaOH) = 0.0510 M (given)
Volume (NaOH solution) = Volume of NaOH added = to be determined
Concentration (CH3COOH) = to be determined (we need to find this)
Let's calculate the concentration of acetic acid (CH3COOH):
Using the equation above:
Concentration (CH3COOH) = (Concentration (NaOH) * Volume (NaOH solution)) / Volume (CH3COOH solution)
We can rearrange the equation to solve for the concentration of acetic acid (CH3COOH):
Concentration (CH3COOH) = (Concentration (NaOH) * Volume (NaOH solution)) / Volume (CH3COOH solution)
Given pH at the equivalence point is 8.51, we know that pOH = 14 - pH = 14 - 8.51 = 5.49.
From this pOH value, we can determine the hydroxide ion concentration ([OH-]) at the equivalence point.
[OH-] = 10^(-pOH) = 10^(-5.49)
Next, we can use the equilibrium constant expression for acetic acid to find its concentration:
Ka = [H3O+][CH3COO-] / [CH3COOH] mymilestonecard
[H3O+] = [OH-] (at equivalence point, the solution is neutral)
Now, we can solve for the concentration of acetic acid (CH3COOH):
[CH3COOH] = [OH-] = 10^(-5.49)
b) Calculate the volume of NaOH added:
Now that we have found the concentration of acetic acid (CH3COOH), we can use the equation:
Volume (NaOH solution) = (Concentration (CH3COOH) * Volume (CH3COOH solution)) / Concentration (NaOH)
Plug in the values:
Volume (NaOH solution) = (Concentration (CH3COOH) * 20.0 mL) / 0.0510 M
Now, we can substitute the value of [CH3COOH] that we found in step (a) to calculate the volume of NaOH added.
- ChenBeier
- Distinguished Member
- Posts: 1569
- Joined: Wed Sep 27, 2017 7:25 am
- Location: Berlin, Germany
Re: Titration
Moles or mass per volumethe formula: Volume (NaOH) = Concentration (NaOH) * Volume (NaOH solution).
Not here because its a weak acid. Valid only for strong acid and base. Like HCl and NaOH. Note pH is 8.51 not 7.H3O+] = [OH-] (at equivalence point, the solution is neutral
pH of salts of weak acid and strong base
pH = ½ (pka + log cHAc + 14)
PLUG IN
8.51 = 0.5 (4,75 + log cHAc + 14)
logcHAc = -1,73
c HAc = 10^(-1,73) = 0,01862 mol/l
The titrated acetate solution contain 0,01862 mol/ l *( 20 ml + V NaOH) = 0,372 mmol HAc in 20 ml acid and x ml NaOH.
This must be equivalent to Hydroxide
Consider Additive volume
0,372 mmol + 0,0186 mol * V NaOH = 0,051 mol/l * V NaOH
0,372 mmol = ( 0,051 mmol/ml - 0,0186 mmol/ml) * V NaOH
V NaOH = 0,372 mmol/0,0324 mmol = 11,48 ml
What means 11,48 ml NaOH was used.