Answer -
Given,
Volume of H3PO4 = 31 ml or 0.031 L [1 ml = 0.001 L]
molarity of NaOH solution = 0.130 M
Volume of NaOH added to reach equivalence point = 27.68 ml or 0.02768 L
Molarity of H3PO4 = ?
H3PO4 (aq) + 3NaOH (aq) → Na3PO4 (aq) + 3H2O (l)
Using Stiochiometry,
It can be analyzed that for 3 moles of NaOH, 1 mole of H3PO3 is required to react completely.
So, moles of H3PO4 = (1/3) * moles of NaOH ---------A
We know that,
Molarity = Moles / Volume in L
Moles = Molarity * Volume in L
So, moles of NaOH in 0.02768 L = 0.130 M * 0.02768 L
moles of NaOH in 0.02768 L = 0.0035984 mol
Using A,
Moles of H3PO4 = (1/3) * 0.0035984 mol = 0.00119946666 mol
Now,
Molarity of H3PO4 = Moles of H3PO4 / Volume of H3PO4
Molarity of H3PO4 = 0.00119946666 mol/ 0.031 L
Molarity of H3PO4 = 0.039 M [Answer]
You may want to reference (Pages 557 - 545) SecuON 0.7 While Compleurig this problem. A...
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