Final answer:
The concentration of the Ca(OH)₂ solution is determined by first finding moles of H₃PO₄ used in the reaction, then applying the stoichiometric ratio to find moles of Ca(OH)₂, and finally dividing by the solution volume. The final concentration is 0.720 M.
Step-by-step explanation:
To calculate the concentration of a Ca(OH)₂ solution when 10.0 ml of 0.600 M H₃PO₄ solution is required to neutralize 12.5 ml of the Ca(OH)₂ solution, we can use a stoichiometry approach. First, write the balanced chemical equation for the reaction:
2 H₃PO₄ (aq) + 3 Ca(OH)₂ (aq) → Ca₃(PO₄)₂ (s) + 6 H₂O (l)
From the balanced reaction, we see that it takes 2 moles of H₃PO₄ to react with 3 moles of Ca(OH)₂. Next, calculate the moles of H₃PO₄ used in the neutralization:
moles H₃PO₄ = 0.600 M × 0.010 L = 0.006 moles
Using the stoichiometric relationship, calculate the moles of Ca(OH)₂:
moles Ca(OH)₂ = (3/2) × moles H₃PO₄ = (3/2) × 0.006 moles = 0.009 moles
Finally, divide the moles of Ca(OH)₂ by the volume of the solution in liters to find the molarity:
Molarity of Ca(OH)₂ = moles Ca(OH)₂ / Volume of the solution (in L) = 0.009 moles / 0.0125 L = 0.720 M
Therefore, the concentration of the Ca(OH)₂ solution is 0.720 M, which corresponds to option C.