1. A buffer solution was prepared in 100 mL volumetric flask by mixing 15 mL of
ID: 1043832 • Letter: 1
Question
1. A buffer solution was prepared in 100 mL volumetric flask by mixing 15 mL of 0.30 M acetic acid (a weak acid) with 15 mL of 0.30 M sodium acetate (a salt of the weak acid) and then diluting with distilled water to the 100 mL mark.
In a similar manner a weak acid solution was made by diluting 15 mL of 0.30 M acetic acid to the 100 mL. (The Ka for acetic acid is 1.76x10-5)
a) Calculate the pH of the buffer solution
I found this answer using the formula pH= pKa + log [ (A-)/(HA-) ]
pH= 4.7
b) Calculate the pH of the weak acid
I found this answer using the ICE table. pH = 2.6
NOW ANSWER THE QUESTION BELOW WHICH IS RELATED TO Q. 1.
Q2. 5.00 mL of 0.010M HCl was added to 20.00 mL of the buffer solution in (1) and also to 20.00 mL of the weak acid solution.
calculate the pH of the weak acid solution (keep in mind that there are two sources of H3O+)
Q3. 5.00 mL of 0.10 M NaOH wa added to 20.00 mL of the buffer solution in (1) and also to 20.00 mL of the weak acid solution.
calculate the pH of the weak acid solution
Explanation / Answer
Hello,
This question takes into consideration behaviour of ions in an aqueous medium. These ions may end up establishing an equilibrium and thus comes the word ionic equilibrium. I am going to solve the given questions while considering the behaviour of Buffer Solutions as well as Common Ion Effect.
Information about the Buffer Solution:
Millimoles of CH3COOH = Molarity x Volume = 0.3 M x 15 ml = 4.5 millimoles
Millimoles of CH3COONa = Molarity x Volume = 0.3 M x 15 ml = 4.5 millimoles
Total volume of Buffer after dilution = 100 ml
pKa of CH3COOH = -log10[Ka] = 4.74
pH of the Buffer solution = 4.74
Information about the Weak Acid (CH3COOH) Solution
Millimoles of CH3COOH = Morality x Volume = 0.3 M x 15 ml = 4.5 millimoles
Volume after dilution = 100 ml
Concentration in Molarity = moles / volume = 4.5 millimoles / 100 ml
pH = -log10[H3O+] = -log10[(Ka.C)0.5] = 3.06
Now, coming to your question
2a) Calculate pH when 5.00 ml of 0.01 M HCl was added to 20 ml of the buffer solution
When HCl is added to the buffer solution, it gives out H+ ions which are taken up by CH3COO- ions to convert into CH3COOH. Thus, the concentration of salt decreases while the concentration of weak acid increases.
Millimoles of HCl added = Molarity x Volume = 0.01 M x 5 ml = 0.05 millimoles
Millimoles of CH3COO- consumed = 0.05 millimoles
Millimoles of CH3COO- left = 4.5 – 0.05 = 4.45 millimoles
Millimoles of CH3COOH formed = 0.05 millimoles
Total millimoles of CH3COOH in the solution = 4.5 + 0.05 = 4.55 millimoles
Since it is the ratio of concentrations, so volume would get cancelled out and we can work out the answer using only the number of moles.
New pH = pH = pKa + log10[salt/acid] = 4.74 + log10[4.45/4.55] = 4.73
2b) Calculate the pH when 5.00 mL of 0.010M HCl was added to 20.00 mL of the weak acid solution.
Now at this stage we need to look into two important aspects of ionic equilibrium. The weak acid is dissociating very feebly in the solution by the following equilibrium:
CH3COOH(aqueous) = CH3COO-(aqueous) + H+(aqueous)
Also, HCl which is being added in the medium dissociates completely in the following manner:
HCl(aqueous) = H+(aqueous) + Cl-(aqueous)
Since HCl is a much stronger acid as compared to CH3COOH so H+ from HCl behaves as the common ion for the CH3COOH dissociation equilibrium. Since in the acetic acid dissociation equilibrium, H+ appears on the product side, so the common ion pushes the equilibrium in the backward direction in accordance to Le-Chatelier’s principle & this further supresses the dissociation of CH3COOH.
Thus [H+]total = [H+]from HCl + [H+]from CH3COOH = [H+]from HCl (since [H+]from CH3COOH is negligible)
Millimoles of HCl added = Molarity x Volume = 0.01 M x 5 ml = 0.05 millimoles
Total volume after addition of HCl = 25 ml
[H+] concentration = 0.05 millimoles / 25 ml
pH = -log10[H+] = -log10[0.002] = 2.7
3a) Calculate pH when 5.00 ml of 0.1 M NaOH was added to 20 ml of the buffer solution
When NaOH is added to the buffer solution, it reacts with CH3COOH ions to convert into CH3COONa . Thus, the concentration of salt increases while the concentration of weak acid decreases.
Millimoles of NaOH added = Molarity x Volume = 0.1 M x 5 ml = 0.5 millimoles
Millimoles of CH3COOH consumed = 0.5 millimoles
Millimoles of CH3COOH left = 4.5 – 0.5 = 4.0 millimoles
Millimoles of CH3COONa formed = 0.5 millimoles
Total millimoles of CH3COONa in the solution = 4.5 + 0.5 = 5.0 millimoles
Since it is the ratio of concentrations, so volume would get cancelled out and we can work out the answer using only the number of moles.
New pH = pH = pKa + log10[salt/acid] = 4.74 + log10[5/4] = 4.83
2b) Calculate the pH when 5.00 mL of 0.10M NaOH was added to 20.00 mL of the weak acid solution.
When you add NaOH to the beaker containing weak acid CH3COOH, an acid base reaction takes place, some weak acid is utilized in the process of consuming all the NaOH and an equivalent amount of salt is formed. The following reaction would thus be established in the titration flask
CH3COOH + NaOH = CH3COONa + H2O
This solution turns out to behave as a buffer solution since some unreacted weak acid would be simultaneously be present with the conjugate base (salt) & the pH of such a solution is calculated using Henderson’s Equation which can be represented as:
pH = pKa + log10[salt/acid]
Millimoles of NaOH added = Molarity x Volume = 0.1 M x 5 ml = 0.5 millimoles
Millimoles of CH3COOH consumed = 0.5 millimoles
Millimoles of CH3COOH left = 0.9 – 0.5 = 0.4 millimoles [100 ml of CH3COOH solution contains 4.5 millimoles of CH3COOH & so, 20 ml of CH3COOH solution would contain 0.9 millimoles of CH3COOH]
Millimoles of CH3COONa formed = 0.5 millimoles
Since it is the ratio of concentrations, so volume would get cancelled out and we can work out the answer using only the number of moles.
New pH = pKa + log10[salt/acid] = 4.74 + log10[0.5/0.4] = 4.83
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