Buffer Preparation a. How many grams of NaH2PO4 . H2O (137.99 g/mole, pKa = 6.86

Question

Buffer Preparation a. How many grams of NaH2PO4 . H2O (137.99 g/mole, pKa = 6.86) are needed to prepare 2.00 L of 0.150 M phosphate buffer b. What volume of 2.00 M NaOH are needed to prepare 2.00 L of 0.150 M phosphate buffer at pH 7.80? c. Write the equilibrium equation for the titration that is occurring. d. You're preparing this buffer by titration with a pH meter so you don't bother to calculate the volume of NaOH needed. You dissolve the phosphate in 1900. mL of distilled water before adjusting the pH. Will you have a problem? If so, what is it? e. Is this an effective buffer at this pH? Why or why not?

Explanation / Answer

(a) We will consider pH 7.8 for the buffer to be prepared,

Usng Hendersen-Haselbalck equation,

pH = pKa + log[base]/[acid])

7.80 = 6.86 + log([base]/[acid])

[base] = 8.71[acid]

[acid] + [base] = 0.15

[base] = 0.15 - [acid]

0.15 - [acid] = 8.71[acid]

[acid] = 0.015 M

[base] = 0.15 - 0.015 = 0.135 M

moles of NaH2PO4.2H2O = [acid] = molarity x volume = 0.015 x 2 = 0.03 mols

grams of NaH2PO4 required = 0.03 x 137.99 = 4.14 g

(b) moles of base = moles of NaOH

moles of base = 0.135 x 2 = 0.27 mols

Volume of NaOH required = 0.27/2 = 0.135 L

(c) The molar concentration of acid and base species would change when you ave dissolved in 1900 ml of water. After adding NaOH the values of acid and base component would be very different from the one you calculated theoretically. So we must have a measure of volume of NaOH used. The pH of the solution would have changed by dilution and thus this will not be an effective buffer anymore.

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