Question

Buffer B Buffer A o.3149 0.0317 Mass of NaC2H3O2 used to prepare buffer (g) 100.0 100.0 Volume of buffer prepared (mL) 1.0 0.1 Molar concentration of HC2H3O2 in buffer (M) 3.80 3.98 Initial pH of buffer Volume of 0.5 M NaOH to raise pH by 2 units (mL) 3. 9mL 34mL 1.5mL Volume of 0.5 M HCI to lower pH by 2 units (mL) 5.5mL Volume of 0.5 M NAOH at equivalence point (mL) ml 36.5mL 1. Buffer capacity has a rather loose definition, yet it is an important property of buffers. A commonly seen definition of buffer capacity is: "The amount of H* or OH that can be neutralized before the pH changes to a significant degree." Use your data to determine the buffer capacity of Buffer A and Buffer B when a strong acid is added to it-that is, how many moles of acid can be added before the buffer is depleted? How many moles of base can be added before the buffer is depleted? Show this work in your "Data and Calculations" section.

Answer 1

Buffer A.

Moles of NaC2H3O2 (a base component of buffer) = 0.031 g/(82 g/mol) = 3.78*10^-4 mol

Moles of HC2H3O2 (an acid component of buffer) = 0.1 mol/L * (100/1000) mL = 0.01 mol

Now, pH = pKa + Log([base]/[acid])

The buffer capacity range is (4.74-1) to (4.74+1) = 3.74 to 5.74

By the addition of acid, pH decreases, i.e. the max. pH for the buffer capacity = 3.74

i.e. 3.74 = 4.74 + Log{(3.78*10^-4 - x)/(0.01+x)}

i.e. (3.78*10^-4 - x)/(0.01+x) = 0.1

i.e. 3.78*10^-4 - x = 0.001 + 0.1x

i.e. 1.1x = 6.22*10^-4

i.e. The buffer capacity = 5.654*10^-4 mol

Buffer B.

Moles of NaC2H3O2 (a base component of buffer) = 0.314 g/(82 g/mol) = 3.83*10^-3 mol

Moles of HC2H3O2 (an acid component of buffer) = 1 mol/L * (100/1000) mL = 0.1 mol

Now, pH = pKa + Log([base]/[acid])

The buffer capacity range is (4.74-1) to (4.74+1) = 3.74 to 5.74

By the addition of acid, pH decreases, i.e. the max. pH for the buffer capacity = 3.74

i.e. 3.74 = 4.74 + Log{(3.83*10^-3 - x)/(0.1+x)}

i.e. (3.83*10^-3 - x)/(0.1+x) = 0.1

i.e. 3.83*10^-3 - x = 0.01 + 0.1x

i.e. 1.1x = 6.17*10^-3

i.e. The buffer capacity = 5.61*10^-3 mol