39 12. Use clean 10 mL graduated cylinders and follow the chart to carefully mea

Question

39 12. Use clean 10 mL graduated cylinders and follow the chart to carefully measure the volume of each reagent indicated into separate, well-drained shell vials. 0.20-M Shell Vial Combined 0.20-M Number Solution acetic acid 0.10-M NaOH H20 HCl acetate 4,42 o mL 3.0 mL 4.0 mL 4.0 mL 3.0 mL 3.0 mL 4.0 mL 4.0 mL 2 2.0 mL ?,17 2.so 7.52 2.0 mL 3.0 mL o mL 2.0 mL 2.0 mL 3.0 mL 2.0 mL 4,92 8 3.0 mL 4.0 mL 3?ay 4,38 ?92-1 3.0 mL 3.0 mL 4.0 mL 2.0 mL 10 11 16. Calculate the initial molarity (after dilution but before any shift to achieve equilibrium) of the acetate and the acetic acid in the combined solution. 17. Calculate the initial moles (due to the combination of solutions or after any neutralization reaction but before any shift to achieve equilibrium) of the acetate ion and the acetic acid present in tubes 1 through 11. In some of the solutions these species come from more than one reagent. In others, acid-base neutralization calculations must be completed before the initial moles can be determined. 18. Determine the ratio of moles of acetate ion to moles of acetic acid for tubes 1 through 11. Express your ratios as 1:1, 1:3, 2:1, etc.

Explanation / Answer

16 and 17. 1 mol = 103 mmol

Shell Vial Number 1. The initial no. of moles of acetic acid = (0.2 mmol/mL * 3 mL) = 0.6 mmol = 6*10-4 mol

The initial molarity of acetic acid = 0.6 mmol/(3+3) mL = 0.1 mmol/mL = 0.1 M

The initial no. of moles of sodium acetate = (0.2 mmol/mL * 3 mL) = 0.6 mmol = 6*10-4 mol

The initial molarity of sodium acetate = 0.6 mmol/(3+3) mL = 0.1 mmol/mL = 0.1 M

Shell Vial Number 2. The initial no. of moles of acetic acid = (0.2 mmol/mL * 4 mL) + (0.2 mmol/mL * 2 mL) = 1.2 mmol = 1.2*10-3 mol

The initial molarity of acetic acid = 1.2 mmol/(4+2) mL = 0.2 mmol/mL = 0.2 M

The initial no. of moles of sodium acetate = (0.2 mmol/mL * 4 mL) = 0.8 mmol = 8*10-4 mol

The initial molarity of sodium acetate = 0.8 mmol/(4+2) mL = 0.133 mmol/mL = 0.133 M

Shell Vial Number 3. The initial no. of moles of acetic acid = (0.2 mmol/mL * 4 mL) = 0.8 mmol = 8*10-4 mol

The initial molarity of acetic acid = 0.8 mmol/(4+2) mL = 0.133 mmol/mL = 0.133 M

The initial no. of moles of sodium acetate = (0.2 mmol/mL * 4 mL) + (0.2 mmol/mL * 2 mL) = 1.2 mmol = 1.2*10-3 mol

The initial molarity of sodium acetate = 1.2 mmol/(4+2) mL = 0.2 mmol/mL = 0.2 M

Shell Vial Number 4. The initial no. of moles of acetic acid = (0.2 mmol/mL * 3 mL) + (0.1 mmol/mL*3 mL) = 0.9 mmol = 9*10-4 mol

The initial molarity of acetic acid = 0.9 mmol/(3+3) mL = 0.15 mmol/mL = 0.15 M

The initial no. of moles of sodium acetate = (0.2 mmol/mL * 3 mL) - (0.1 mmol/mL*3 mL) = 0.3 mmol = 3*10-4 mol

The initial molarity of sodium acetate = 0.3 mmol/(3+3) mL = 0.05 mmol/mL = 0.05 M

18. According to Henderson-Hasselbulch equation: pH = pKa + Log(nsodium acetate/nacetic acid)

Shell Vial Number 1. 4.42 = 4.74 + Log(nsodium acetate/nacetic acid)

i.e. Log(nsodium acetate/nacetic acid) = 4.42-4.74 = -0.32

i.e. nsodium acetate/nacetic acid = 10-0.32 = 0.5 = 1:2

Shell Vial Number 2. 4.19 = 4.74 + Log(nsodium acetate/nacetic acid)

i.e. Log(nsodium acetate/nacetic acid) = 4.19-4.74 = -0.55

i.e. nsodium acetate/nacetic acid = 10-0.55 = 0.3 = 1:3

Shell Vial Number 3. 4.77 = 4.74 + Log(nsodium acetate/nacetic acid)

i.e. Log(nsodium acetate/nacetic acid) = 4.77-4.74 = 0.03

i.e. nsodium acetate/nacetic acid = 100.03 ~ 1 = 1:1

Shell Vial Number 4. 2.59 = 4.74 + Log(nsodium acetate/nacetic acid)

i.e. Log(nsodium acetate/nacetic acid) = 2.59-4.74 = -2.15

i.e. nsodium acetate/nacetic acid = 10-2.15 ~ 0.01 = 1:100

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