Assume that the reaction used to prepare the buffer goes to completion (i.e., th

Question

Assume that the reaction used to prepare the buffer goes to completion (i.e., that all the strong base reacts) and use the data in Table A to calculate the initial concentration of the weak acid and its conjugate base in each of the three buffer solutions.

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This experiment is done by students working in groups of three. During Week One, each group will be assigned one specific weak acid to work with: acetic acid (CH3CO2H), formic acid (HCO2H), or propanoic acid (CH3CH2CO2H). Each group will prepare three buffer solutions by mixing their assigned weak acid with different amounts of NaOH, as indicated in Table A.

TABLE A

mL of 0.50 M weak acid

mL of H2O

Buffer

mL of 0.50 M weak acid

mL of 0.25 M NaOH

mL of H2O

1 50.00 50.00 0.00 2 50.00 35.00 15.00 3 50.00 25.00 25.00

Explanation / Answer

Buffer 1

Moles of acid = Molarity of weak acid x Volume of acid in L

50 mL = 0.50L

Moles of acid = 0.50 M x 0.050L = 0.025 moles

Moles of NaOH = Molarity of NaOH x Volume of NaOH in L

Moles of NaOH = 0.25M x 0.050L = 0.0125 moles

Final volume of solution = 50ml + 50 ml = 100 ml = 0.1 L

Initial concentration of weak acid = 0.025 moles/0.1L = 0.250M

Initial concentration of NaOH = 0.0125 moles/0.1L = 0.125M

Buffer 2

Moles of acid = 0.50 M x 0.050L = 0.025 moles

Moles of NaOH = Molarity of NaOH x Volume of NaOH in L

Moles of NaOH = 0.25M x 0.035L = 0.0085 moles

Final volume of solution = 50ml acid + 35 ml NaOH + 15 ml Water = 100 ml = 0.1 L

Initial concentration of weak acid = 0.025 moles/0.1L = 0.250M

Initial concentration of NaOH = 0.0085 moles/0.1L = 0.0850M

Buffer 3

Moles of acid = 0.50 M x 0.050L = 0.025 moles

Moles of NaOH = Molarity of NaOH x Volume of NaOH in L

Moles of NaOH = 0.25M x 0.025L = 0.00625 moles

Final volume of solution = 50ml acid + 25 ml NaOH + 25 ml Water = 100 ml = 0.1 L

Initial concentration of weak acid = 0.025 moles/0.1L = 0.250M

Initial concentration of NaOH = 0.00625 moles/0.1L = 0.0625M

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