Answer:
0.4058L of 1.0M H3PO4
0.2192L of 1.5M NaOH
Explanation:
The pKa of the H3PO4 / H2PO4- buffer is 2.12
To solve this question we must use H-H equation for this system:
pH = pKa + log [H2PO4-] / [H3PO4]
2.75 = 2.12 + log [H2PO4-] / [H3PO4]
0.63 = log [H2PO4-] / [H3PO4]
<em>4.2658 = [H2PO4-] / [H3PO4] (1)</em>
Where [] could be taken as the moles of each reactant
As you have H3PO4 solution, the reaction with NaOH is:
H3PO4 + NaOH → H2PO4- + Na+ + H2O
As you can see, both H3PO4 and H2PO4- comes from the same 1.0M H3PO4 solution
The moles of H3PO4 are:
[H3PO4] = Moles H3PO4 - Moles NaOH
And for H2PO4-:
[H2PO4-] = Moles NaOH added
Replacing in (1):
4.2658 = [Moles NaOH] / [Moles H3PO4 - Moles NaOH]
4.2658 Moles H3PO4 - 4.2658 moles NaOH = Moles NaOH
4.2658 Moles H3PO4 = 5.2658 moles NaOH <em>(1)</em>
<em></em>
In volume:
0.625L = Moles H3PO4 / 1.0M + Moles NaOH / 1.5M
0.625 = Mol H3PO4 + 0.6667 Moles NaOH <em>(2)</em>
Replacing (2) in (1):
4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4 / 0.6667)
4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4) / 0.6667
4.2658 Moles H3PO4 = 5.2658*(0.9375 - 1.5 mol H3PO4)
4.2658 Moles H3PO4 = 4.9367 -7.8983 mol H3PO4
12.1641 mol H3PO4 = 4.9367
Mol H3PO4 = 0.4058moles * (1L / 1.0moles) =
0.4058L of 1.0M H3PO4
And:
0.625L - 0.4058L =
0.2192L of 1.5M NaOH
