Answer:
3mL of 5M NaOH must be added to adjust the pH to 7.20
Explanation:
When NaOH is added to phosphoric acid, H₃PO₄, the reaction that occurs are:
NaOH + H₃PO₄ ⇄ NaH₂PO₄ + H₂O pKa1 = 2.15
NaOH + NaH₂PO₄ ⇄ Na₂HPO₄ + H₂O pKa2 = 7.20
NaOH + Na₂HPO₄ ⇄ Na₃PO₄ + H₂O pKa3 = 12.38
We can adjust the pH at 7.20 = pKa2 if NaH₂PO₄ = Na₂HPO₄. To make that, we must convert, as first, all H₃PO₄ to NaH₂PO₄ and the half of NaH₂PO₄ to Na₂HPO₄. To solve this question we need to find the moles of phophoric acid in the initial solution. 1.5 times these moles are the moles of NaOH that must be added to fix the pH to 7.20:
Moles H₃PO₄:
100mL = 0.100L * (0.100mol / L) = 0.0100 moles H₃PO₄
Moles NaOH:
0.0100 moles H₃PO₄ * 1.5 = 0.0150 moles NaOH
Volume NaOH:
0.0150 moles NaOH * (1L / 5moles) = 3x10⁻³L 5M NaOH are required =
3 mL of 5 Molar NaOH is required to adjust the pH of phosphoric acid.
It is the negative log of the concentration of Hydrogen ions in the solution.
To calculate the volume of NaOH first, calculate the moles of NaOH and H₃PO₄.
Moles of H₃PO₄.
[tex]\rm moles \ of \ H_3PO_4 = 100\rm \ mL = 0.100\rm \ L \times (0.100 \ mol / L)\\\\\rm moles \ of \ H_3PO_4 = 0.01[/tex]
The moles of NaOH:
[tex]\rm Moles \ of \ NaOH =0.01 \ moles \ H_3PO_4\times 1.5 \\\\\rm Moles \ of \ NaOH= 0.0150[/tex]
The volume of NaOH:
[tex]\rm Volume\ of \ NaOH = \rm 0.0150\ moles\ NaOH \times (1 \ L / 5 \ moles) \\\\\rm Volume\ of \ NaOH = 3\times 10^{-3} L[/tex]
Therefore, 3 mL of 5 Molar NaOH is required to adjust the pH of phosphoric acid.
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