Final answer:
To determine the number of milliliters of 0.00300 M phosphoric acid required to neutralize 35.00 mL of 0.00150 M calcium hydroxide, we can use the stoichiometry of the balanced equation. The balanced equation is 2 H3PO4(aq) + 3 Ca(OH)2(aq) → Ca3(PO4)2(aq) + 6 H2O(l). Calculating the moles and using the concentration, we find that it would require approximately 0.105 mL of 0.00300 M phosphoric acid to neutralize 35.00 mL of 0.00150 M calcium hydroxide.
Step-by-step explanation:
The balanced chemical equation for the reaction is:
2 H3PO4(aq) + 3 Ca(OH)2(aq) → Ca3(PO4)2(aq) + 6 H2O(l)
To determine the number of milliliters of phosphoric acid (H3PO4) required to neutralize 35.00 mL of calcium hydroxide (Ca(OH)2), we need to use the stoichiometry of the balanced equation.
From the balanced equation, we can see that 2 moles of H3PO4 react with 3 moles of Ca(OH)2. First, we need to find the number of moles of Ca(OH)2 in 35.00 mL of 0.00150 M Ca(OH)2:
moles of Ca(OH)2 = volume (L) × concentration (M)
moles of Ca(OH)2 = 0.03500 L × 0.00150 M
moles of Ca(OH)2 = 0.0000525 mol
Now, we can use the stoichiometry of the balanced equation to find the number of moles of H3PO4:
moles of H3PO4 = (0.0000525 mol Ca(OH)2) × (2 mol H3PO4 / 3 mol Ca(OH)2)
moles of H3PO4 = 0.000035 mol
Finally, we can convert the moles of H3PO4 to milliliters using its concentration:
volume (mL) = moles × concentration × conversion factor
volume (mL) = 0.000035 mol × 0.00300 M × (1000 mL / 1 L)
volume (mL) = 0.105 mL
Therefore, it would require approximately 0.105 mL of 0.00300 M phosphoric acid to neutralize 35.00 mL of 0.00150 M calcium hydroxide.