Add approximately 4 grams of sodium acetate to 4 mL of 6 M acetic acid. Dilute w

Question

Add approximately 4 grams of sodium acetate to 4 mL of 6 M acetic acid. Dilute with distilled water to a final volume of 40 mL and agitate the solution until the sodium acetate has completely dissolved. Label six clean test tubes A through F. Place 10 mL of distilled water into test tube A and test tube D. Place 10 mL of the sodium acetate buffer prepared above to test tubes B, C, K. and F. To test tubes A, B, and C add 2 drops of phenolphthalein indicator. To test tubes D, E, and F add 2 drops of methyl orange indicator. Place a cork or your thumb over the top of each test tube and invert several times to mix. Record the color of each solution on the data sheet. Next we will study the effect of the addition of moderate amounts of acid or base to these solutions by adding 2 drops of 3 M NaOH to tubes A and B and by adding 2 drops of 3 M HCl to tubes D and B. Mix the solutions by inverting the test tubes as done previously. Record the color of each solution on the data sheet. Determine how many drops of 3 M NaOH must be added to test tube C before it has the same color as test tube A. Record this value on the data sheet. Determine and then record how many drops of 3 M HCl must be added to test tube F before it has the same color as test tube D.

Explanation / Answer

Buffer solution

buffer system : CH3COOH/CH3COONa

initial concentration of [CH3COOH] in buffer solution = 6 M x 4 ml/40 ml = 0.60 M

initial concentration of [CH3COONa] in buffer solution = 4/82.03 x 0.040 = 1.22 M

initial pH of buffer solution,

pKa of acetic acid = 4.75

pH = pKa + log(base/acid)

      = 4.75 + log(1.22/0.60)

      = 5.06

1 drop = 0.05 ml

For test tube A, after 2 drops (0.03 ml) of 3 M NaOH was added

[OH-] = 3 M x 0.03 ml/10.03 ml = 0.009 M

pOH = -log[OH-] = 2.04

pH = 14 - pOH = 11.95

For test tube B, after 2 drops (0.03 ml) of 3 M NaOH was added

pH = 4.75 + log[(1.22 M x 10 ml + 0.09)/(0.6 M x 10 ml - 0.09)]

     = 5.07

For test tube D, after 2 drops (0.03 ml) of 3 M HCl was added

[H+] concentration = 3 M x 0.03 ml/10.03 ml = 0.009 M

pH = -log[H+] = 2.04

For test tube E, after 2 drops (0.03 ml) of 3 M HCl was added

pH = 4.75 + log[(1.22 M x 10 ml - 0.09)/(0.6 M x 10 ml + 0.09)]

     = 5.05

let x moles of 3 M NaOH was added to test tube C to reach pH of test tube A,

11.95 = 4.75 + log(1.22 M x 10 ml + x)/(0.6 M x 10 ml - x)]

9.5 x 10^7 - 1.6 x 10^7x = 12.2 + x

x = 5.94 mmol

volume of NaOH to be added = 5.94 mmol/3 M = 1.98 ml = 40 drops

let x moles of 3 M HCl was added to test tube F to reach pH of test tube D,

2.04 = 4.75 + log(1.22 M x 10 ml - x)/(0.6 M x 10 ml + x)]

0.0117 + 0.002x = 12.2 - x

x = 12.16 mmol

volume of HCl to be added = 12.16 mmol/3 M = 4.05 ml = 81 drops

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