What volume in mL of 0.0985 M sodium hydroxide solution is required to reach the equivalence point in the complete titration of a 15 mL sample of 0.124 M phosphoric acid?

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To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation between sodium hydroxide (NaOH) and phosphoric acid (H3PO4).

The balanced chemical equation for the reaction between NaOH and H3PO4 is:
3 NaOH + H3PO4 → Na3PO4 + 3 H2O

From the balanced equation, we can see that the ratio between NaOH and H3PO4 is 3:1. This means that for every 3 moles of NaOH, we need 1 mole of H3PO4 to react completely.

Let's start by finding the moles of H3PO4 in the 15 mL sample of 0.124 M phosphoric acid:
Moles of H3PO4 = volume (in L) × concentration (in mol/L)
Moles of H3PO4 = 15 mL × (1 L / 1000 mL) × 0.124 mol/L
Moles of H3PO4 = 0.00186 mol

Now, using the stoichiometric ratio, we can determine the moles of NaOH needed to react completely with the H3PO4:
Moles of NaOH = (1/3) × moles of H3PO4
Moles of NaOH = (1/3) × 0.00186 mol
Moles of NaOH = 0.00062 mol

Finally, we can calculate the volume of the 0.0985 M NaOH solution needed:
Volume of NaOH (in L) = moles of NaOH / concentration (in mol/L)
Volume of NaOH (in L) = 0.00062 mol / 0.0985 mol/L
Volume of NaOH (in L) = 0.00630 L

To convert the volume to milliliters (mL):
Volume of NaOH (in mL) = 0.00630 L × 1000 mL/L
Volume of NaOH (in mL) = 6.3 mL

Therefore, approximately 6.3 mL of 0.0985 M sodium hydroxide (NaOH) solution is required to reach the equivalence point in the complete titration of a 15 mL sample of 0.124 M phosphoric acid.