1a. The balanced equation for the reaction is:
1b. The number of mole of Ca(OH)₂ is 0.0247 mole
1c. The number of mole of H₃PO₄ is 0.0165 mole.
1d. The concentration of H₃PO₄ is 0.432 mol/L
2. The new concentration of the H₃PO₄ solution is 0.0432 mol/L
3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O
Volume of Ca(OH)₂ = 71 mL = 71 / 1000 = 0.071 L
Concentration of Ca(OH)₂ = 0.348 mol/L
Mole = Concentration × Volume
Mole = 0.348 × 0.071
3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
3 moles of Ca(OH)₂ reacted with 2 moles of H₃PO₄.
Therefore,
0.0247 moles of Ca(OH)₂ will react with = [tex]\frac{0.0247 * 2}{3}[/tex] = 0.0165 mole of H₃PO₄.
Thus, the number of mole of H₃PO₄ is 0.0165 mole
Volume of H₃PO₄ = 38.20 mL = 38.20/ 1000 = 0.0382 L
Mole of H₃PO₄ = 0.0165 mole
[tex]Concentration = \frac{mole}{volume} \\\\Concentration = \frac{0.0165}{0.0382}[/tex]
Initial Volume (V₁) = 10 mL
Initial concentration (C₁) = 0.432 mol/L
New volume (V₂) = 100 mL
The new concentration of the H₃PO₄ solution can be obtained as follow:
0.432 × 10 = C₂ × 100
4.32 = C₂ × 100
Divide both side by 100
C₂ = [tex]\frac{4.32}{100}\\[/tex]
Therefore, the new concentration of the H₃PO₄ solution is 0.0432 mol/L
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