A 15.4 mL aliquot of 0.204 MH3PO4(aq) is to be titrated with 0.17 MNaOH(aq). What volume (mL) of base will it take to reach the equivalence point?

55.44L of the 0.17M NaOH are required

Explanation:

Phosphoric acid, H3PO4, reacts with NaOH as follows:

H3PO4 + 3NaOH → Na3PO4 + 3H2O

Where 1 mole of H3PO4 reacts with 3 moles of NaOH

To solve this question we need to find the moles of H3PO4 in the aliquot. Using the balanced equation we can find the moles of NaOH and its volume with the concentration (0.17M) as follows:

Moles H3PO4:

15.4mL = 0.0154L * (0.204mol/L) = 0.00314 moles H3PO4

Moles NaOH:

0.00314 moles H3PO4 * (3mol NaOH / 1mol H3PO4) = 0.009425moles NaOH

Volume NaOH:

0.009425moles NaOH * (1L/0.17moles NaOH) = 0.05544L 0.17M NaOH =

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