Suppose there is 1 .00 L of an aqueous buffer containing 60.0 mmol of acetic acid…
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Question “Suppose there is 1 .00 L of an aqueous buffer containing 60.0 mmol of acetic acid…”
Answer
Volume of the solution: 1.00 L
The initial concentrations of acetic acid ions and acetate are
You can write the following description of the acid dissociation:
To determine equilibrium concentrations, you can make an ICE table.
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|
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Initial, M | 0.0600 | 0.0400 | 0 |
Change is M | -x | +x | +x |
Equilibrium and M | 0.0600 -x | 0.0400+x | x |
The acid’s pKa is 4.76
hence,
Hence,
Hence, the pH of the solution is 4.58.
Alternatively, you can directly use
Henderson-Hasselbalch equation
The solution must be pH adjusted by adding NaOH. The addition of NaOH to the solution can be represented as
The target pH can be used to calculate the ratio of acid and base.
As the equation below indicates, adding NaOH causes an increase in CH3COO-base concentration and decreases in acid CH3COOH concentration.
We can thus create a new ICE Table
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Initial, mol | 0.0600 -x | Y | 0.0400+x |
Mole, change | -y | -y | +y |
Equilibrium and mol | 0.06-x-y | 0 | 0.0400+x+y |
You will need to add y mole of NaOH to react with the weak acid to make conjugate base equal in mol y.
We can ignore it because x is much less than 0.06 or 0.04.
The equilibrium moles for acid and base have been renamed.
Therefore, use the above-mentioned ratio of acid to base
To raise the pH to 4.93, we will need to add 0.0277 mol NaOH.
Concentration of NaOH solution is available at 2.00 M = 2.00 Mol/L
Therefore, the volume that contains 0.0297 mol NaOH (0.0197 mol) is
Hence, the required volume of NaOH to be added is
approximately 9.85 mL.
Conclusion
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