ANSWER B,C & DThe Ka value of acetic (ethanoic) acid is 1.75x 10-5 .a) Calculate the pH of a 0.10M solution of CH3COOH(aq).b) 50mL of 0.10M of acetic acid and 50mL of 0.10M of sodium acetate are mixed. What is the pH of the resulting solution? Include any assumptions made.c) 0.0025 mole of NaOH is added to 100mL of the solution in b). What is now the pH of the resulting solution? Assume that volume of the solution does not change when NaOH is added.d) Comment on the effectiveness of the solution in b) as a buffer giving a reason. How could a more effective buffer solution be prepared?

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Asked Oct 23, 2019
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ANSWER B,C & D
The Ka value of acetic (ethanoic) acid is 1.75x 10-5 .

a) Calculate the pH of a 0.10M solution of CH3COOH(aq).

b) 50mL of 0.10M of acetic acid and 50mL of 0.10M of sodium acetate are mixed. What is the pH of the resulting solution? Include any assumptions made.

c) 0.0025 mole of NaOH is added to 100mL of the solution in b). What is now the pH of the resulting solution? Assume that volume of the solution does not change when NaOH is added.

d) Comment on the effectiveness of the solution in b) as a buffer giving a reason. How could a more effective buffer solution be prepared?

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Expert Answer

Step 1

According to the Henderson equation,

 

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pH=pK, +log [salt] [acid] pK-logK -log(1.75x10 a 4.75 [0.10] pH 4.75+log [0.10] =4.75+0 =4.75

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Step 2

When 0.0025 moles of NaOH are added to the 100ml of the solution

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no. of moles molarity= volume 0.0025 0.1 -0.025M

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Step 3

This basically adds 0.025M of Na ions and it decreases the acid...

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[salt] pH-pKlog [acid] [0.10+0.025 4.75+log [0.10-0.025] -4.75 0.2218 =4.8718

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