Question

You may want to reference (Pages 557 - 545) SecuON 0.7 While Compleurig this problem. A 31,00 mL sample of an unknown H3PO, solution is titrated with a 0.130 M NaOH solution. The equivalence point is reached when 27.68 mL of NaOH solution is added

Answer 1

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]