Question

describe how you would prepare exactly 100 mL of .100 M picolinate buffer, ph 5.50. possible starting materials are pure picolinic acid (FW=123.11), 1.0 M HCl, and 1 M NaOH. Approximately how many milliliters of HCl and NaOH will be required?

Answer 1

The required pH = 5.5

According to the Hendersen-Hasselbalch equation, you can write as shown below.

pH = pKa + Log([picolinate]/[picolinic acid])

i.e. 5.5 = 5.4 + Log(x mL * 1 M / y), where x = no. of milliliters of NaOH required, y = no. of mmol of picolinic acid

i.e. Log(x / y) = 5.5 - 5.4 = 0.1

i.e. x / y = 10^0.1 = 1.25, i.e. x = 1.25 y ........... equation 1

x + y = 100 mL * 0.1 mmol/mL = 10 mmol .... equation 2

From equations 1 and 2, you can obtain as follows.

1.25 y + y = 10

i.e. y = 10/2.25 = 4.4

Now, you have to take 4.4 moles of picolinic acid, i.e. 4.4 mol * 123.11 g/mol = 0.54 g of picolinic acid

From equation 1, x = 1.25 * 4.4 = 5.5

i.e. 5.5 mL of 1 M NaOH is required to added to picolinic acid.