Question

Phosphoric acid has three titratable hydrogen ions; the respective equilibria are expressed as follows:

H₃PO₄ --> /<-- H⁺ + H₂PO₄^1-                                     pK₁ = 2.1

H₂PO₄^1- -->/<-- H⁺ + HPO₄^2-                          pK₂ = 6.8

HPO₄^2- -->/<-- H⁺ + PO₄^3-                             pK₃ = 12.4

How would you prepare 1L of a 1 M phosphate buffer, pH 7.5? Your available chemicals include solid NaH₂PO₄, solid Na₂HPO₄, solid NaOH, and 6 M HCl solution. Indicate how you would prepare this buffer.

Answer 1

Buffer solution:

Buffer solution is made by the addition of acid and base solution. The pH of this solution is remain unchanged by adding acid or base.   In general the pH of this solution is constant.

if we want the pH of your buffer to be 7.5 , then use the pKa of 6.8:

pH = pKa + log ([Base]/[Acid])

7.5 = 6.8 + log ([Base]/[Acid])

log ([Base]/[Acid])=0.7

([Base]/[Acid]) =10^0.7

ratio of [Base]/[Acid] = 5.012

and [Base] = 5.012 [Acid]

The molarity of the buffer is the sum of the molarities of the acid and conjugate base or the sum of [Acid] + [Base]. For a 1 M buffer (selected to make the calculation easy),

[Acid] + [Base] = 1

[Acid] + 5.012 [Acid] =1

6.012 [Acid] = 1

[Acid]= 0.166

And

[Base] = 1 - [Acid]

= 1 – 0.166

= 0.834 moles/L

Here volume is also 1.0 L thus number of mole s= molarity

Prepare the solution by mixing 0.166 moles of monosodium phosphate and 0.834 moles of disodium phosphate in a little less than a liter of water.

Or

0.166 moles of monosodium phosphate *119.98 g/mol = 19.91668 g

0.834 moles of disodium phosphate * 141.96 g/ mole= 118.39 g