Final answer:
To calculate the volume of 2.3 M NaOH needed to adjust pH from 4.0 to 9.0 in a solution of phosphoric acid, perform stoichiometric calculations using the molar amounts of phosphoric acid and NaOH, bearing in mind the relevant pKa value and the Henderson-Hasselbalch equation.
Step-by-step explanation:
To estimate the volume of a 2.3 M NaOH solution needed to adjust the pH from 4.0 to 9.0 in 61 ml of a 68 mM phosphoric acid solution, we must first recognize that the pKa values of phosphoric acid are: pKa1= 2.15, pKa2=7.20, and pKa3=12.35. Because we are adjusting the pH to 9.0, which is between pKa2 and pKa3, we can assume that the second dissociation is the relevant one for our calculation.
Using the Henderson-Hasselbalch equation for the pH at the second dissociation:
pH = pKa + log([A-]/[HA])
Here, [A-] represents the conjugate base concentration and [HA] represents the conjugate acid concentration. We want to adjust the pH to 9.0, and at this point, [A-] will approximately equal [HA].
For the neutralization, the molar amount of NaOH needed equals the molar amount of H2PO4-. Here, the molar amount of phosphoric acid is:
Moles of H2PO4- = 61 ml × 0.068 mmol/ml × 1 mol/1000 mmol
The concentration of NaOH is given as 2.3 M, and using the molarity concept:
Volume of NaOH = Moles of NaOH / Molarity of NaOH.
Assuming complete neutralization, moles of NaOH needed will be equal to moles of H2PO4-. So, the final calculation for the volume of NaOH required can be determined using the above relationship.