Answer:
15000 mL or 15 mL.
Step-by-step explanation:
To determine the volume of 0.00300 M phosphoric acid required to neutralize 45.00 mL of 0.00150 M calcium hydroxide, we can use the balanced chemical equation provided:
3 Ca(OH)2(aq) + 2 H₃PO₄(aq) → 1 Ca₃(PO₄)2(aq) + 6 H₂O(l)
From the balanced equation, we can see that the stoichiometric ratio between calcium hydroxide (Ca(OH)2) and phosphoric acid (H₃PO₄) is 3:2. This means that for every 3 moles of calcium hydroxide, we need 2 moles of phosphoric acid to react completely.
Now let's calculate the moles of calcium hydroxide in 45.00 mL of 0.00150 M solution:
moles of Ca(OH)2 = volume (L) × concentration (mol/L)
moles of Ca(OH)2 = 45.00 mL × (1 L / 1000 mL) × 0.00150 mol/L
moles of Ca(OH)2 = 0.0675 mol
Since the stoichiometric ratio between calcium hydroxide and phosphoric acid is 3:2, we can use this ratio to find the moles of phosphoric acid required:
moles of H₃PO₄ = (2/3) × moles of Ca(OH)2
moles of H₃PO₄ = (2/3) × 0.0675 mol
moles of H₃PO₄ = 0.045 mol
Now, let's calculate the volume of 0.00300 M phosphoric acid needed to supply 0.045 mol:
volume (L) = moles / concentration (mol/L)
volume (L) = 0.045 mol / 0.00300 mol/L
volume (L) = 15 L
Since we have the volume in liters, we can convert it to milliliters:
volume (mL) = volume (L) × (1000 mL / 1 L)
volume (mL) = 15 L × 1000 mL
volume (mL) = 15000 mL
Therefore, the volume of 0.00300 M phosphoric acid required to neutralize 45.00 mL of 0.00150 M calcium hydroxide is 15000 mL or 15 mL.