Solution #1 50.0 mL of 0.0750 M acetic acid and 0.308 grams of sodium acetate So

ID: 1019656 • Letter: S

Question

Solution #1

50.0 mL of 0.0750 M acetic acid and 0.308 grams of sodium acetate

Solution #2

25.0 mL of 0.0750 M acetic acid, 25.0 mL of deionized water, and 0.308 grams of sodium acetate

Solution #3

50.0 mL of 0.0750 M acetic acid and 0.154 grams of sodium acetate

Solution #4

50.0 mL of 0.0750 M acetic acid

Solution #5

50.0 mL deionized water and 0.308 grams of sodium acetate

Solution #1

50.0 mL of 0.0750 M acetic acid and 0.308 grams of sodium acetate

Solution #2

25.0 mL of 0.0750 M acetic acid, 25.0 mL of deionized water, and 0.308 grams of sodium acetate

Solution #3

50.0 mL of 0.0750 M acetic acid and 0.154 grams of sodium acetate

Solution #4

50.0 mL of 0.0750 M acetic acid

Solution #5

50.0 mL deionized water and 0.308 grams of sodium acetate

Solution #1 : concentration of sodium acetate in 50 ml buffer = 0.308 g/82.0343 g/mol x 0.05 L = 0.075 M

Volume of HCl required for reaction = 0.075 M x 25 ml/0.1 M = 18.75 ml

Solution #1 : Volume of NaOH required for reaction = 0.075 M x 25 ml/0.1 M = 18.75 ml

Solution #2 : concentration of sodium acetate in 50 ml buffer = 0.308 g/82.0343 g/mol x 0.05 L = 0.075 M

Volume of HCl required for reaction = 0.075 M x 25 ml/0.1 M = 18.75 ml

Solution #2 : Volume of NaOH required for reaction = 0.075 M x 25 ml/0.1 M = 18.75 ml

Solution #3 : concentration of sodium acetate in 50 ml buffer = 0.154 g/82.0343 g/mol x 0.05 L = 0.0375 M

Volume of HCl required for reaction = 0.0375 M x 25 ml/0.1 M = 9.375 ml

Solution #3 : Volume of NaOH required for reaction = 0.075 M x 25 ml/0.1 M = 18.75 ml

Solution #4 : concentration of sodium acetate in 50 ml buffer = 0 M

Volume of HCl required for reaction = 0 ml

Solution #4 : Volume of NaOH required for reaction = 0.075 M x 25 ml/0.1 M = 18.75 ml

Solution #5 : concentration of sodium acetate in 50 ml buffer = 0.308 g/82.0343 g/mol x 0.05 L = 0.075 M

Volume of HCl required for reaction = 0.075 M x 25 ml/0.1 M = 18.75 ml

Solution #5 : Volume of NaOH required for reaction = 0 ml

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