Question

A 0.100 g sample of phosphoric acid (H3PO4, MW 99.0 g/mol), a tri-protic acid, requires 35.0 mL of a NaOH solution for neutralization. What is the molarity of the NaOH solution

Answer 1

Answer:

0.0866 M

Explanation:

answer key

Answer 2

The molarity of the NaOH solution has been 0.0865 M.

The neutralization reaction of the Phosphoric acid and NaOH results in the formation of sodium phosphate and water.

The balanced chemical equation can be given as;

$\rm H_3PO_4\;+\;3\;NaOH\;\rightarrow\;Na_3PO_4\;+\;3\;H_2O$

Thus for the neutralization of 1 mole of Phosphoric acid, 3 moles of NaOH has been used.

The moles of phosphoric acid in 0.100 grams sample:

Moles = $\rm \dfrac{weight}{molecular\;weight}$

Moles of phosphoric acid = $\rm \dfrac{0.100}{99}$

Moles of phosphoric acid = 0.001 mol

For the neutralization of 0.001 moles of NaOH:

1-mole Phosphoric acid = 3 moles NaOH

0.001-mole Phosphoric acid =0.003 moles of NaOH.

The molarity of the NaOH solution:

Molarity = $\rm \dfrac{moles}{Volume\;(L)}$

Molarity of NaOH = $\rm \dfrac{0.003}{0.035}$

Molarity of NaOH = 0.0865 M

The molarity of the NaOH solution has been 0.0865 M.

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