Discuss in detail the theory behind the buffer you created and the amount of aci

Question

Discuss in detail the theory behind the buffer you created and the amount of acid and base you added. Perform the appropriate calculation to determine the theoretical pH of the buffer solution after 10 drops of 0.1M HCl was added.   

Ka of acetic acid = 1.8 x 10-5

To start this calculation, find the equilibrium concentrations of the HA <=> H+ + A- for the buffer you set up.

Then shift that equilibrium by adding 10 drops of 0.1M HCl.

**This tends to through students for a loop. Perhaps if we phrase it this way. A buffer was established with 2.0mL of 0.833M acetic acid. 0.5mL of 0.1M HCl is added. What is the new pH of the buffer?

Discuss in a complete sentence how your theoretical calculation here is the same or different than what was obtained above.

REMEMBER - These lab forms take the place of a formal report. Your discussion shows that you understand what you saw and can link the module's topics to this application.

The buffer was prepared by adding 4 drops of 6M NaOH in 2 mL of 0.833 M acetic acid. How do we find the concentration of sodium acetate?

Explanation / Answer

Acetic acid is weak acid. So, it dissociates partially, and solution contains both acetic acid and acetate.

Since acetic acid is weak acid and acetate is its conjugate base, the mixture of acetic acid and acetate is a buffer.

Suppose 0.833 M acetic acid dissociation is x M.

CH3COOH <------>   CH3COO- + H+

K = [CH3COO- ][H+ ] / [CH3COOH]

1.8x10-5 = x2 / (0.833-x)

1.8x10-5 = x2 / (0.833)

Solving we get x = 0.00387

So, Initially 0.833 M acetic acid solution contains 0.00387 M CH3COO- and 0.829 M CH3COOH.

The HCl added is react with CH3COO-

CH3COO- + H+ ------> CH3COOH

So, added amount of HCl subtract the initial amount of CH3COO- and add to the initial amount of CH3COOH.

Initial amount of CH3COO- = 2.0 mL x 0.00387 M = 0.00774 mmol

Initial amount of CH3COOH = 2.0 mL x 0.829 M = 1.658 mmol

Added amount of HCl = 0.5 mL x 0.1 M = 0.05 mmol

Here, mmol HCl is greater tha mmol CH3COO- So, all the amount of CH3COO- is converted in to CH3COOH

Solution is not a buffer.

Final amount of HCl = 0.05 mmol - 0.00774 mmol = 0.04226 mmol

Since HCl is strong acid,

[HCl] = [H+ ] = (0.04226 mmol)/(2.0 mL + 0.5 mL) = 0.0169 M

pH =-log[H+ ] = -log(0.0169) = 1.77

NaOH reacts with CH3COOH.

NaOH + CH3COOH -----> CH3COONa + H2O

So, added amount of NaOH is equal to sodium acetate

Concentration of sodium acetate =( volume of NaOH x Molarity of NaOH)/ total voluem of solution.

Total volume of solution is volume of NaOH and volume of CH3COOH.

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