calculate the volume of 1.07M phosphoric acid solution necessary to react with 22.9 ml of 0.150M Mg(OH)
To calculate the volume of 1.07M phosphoric acid solution necessary to react with 22.9 ml of 0.150M Mg(OH)₂, we need to use the balanced chemical equation and the concept of stoichiometry.
First, let's write the balanced chemical equation for the reaction between phosphoric acid (H₃PO₄) and magnesium hydroxide (Mg(OH)₂):
2 H₃PO₄ + 3 Mg(OH)₂ → Mg₃(PO₄)₂ + 6 H₂O
From the balanced equation, we can see that the stoichiometric ratio between H₃PO₄ and Mg(OH)₂ is 2:3. This means that for every 2 moles of H₃PO₄, we need 3 moles of Mg(OH)₂.
Now, let's calculate the number of moles of Mg(OH)₂ in 22.9 ml of 0.150M Mg(OH)₂ solution:
moles of Mg(OH)₂ = volume (in L) × concentration (in M)
moles of Mg(OH)₂ = 0.0229 L × 0.150 M
moles of Mg(OH)₂ = 0.003435 moles
Using the stoichiometry of the balanced equation, we can determine the number of moles of phosphoric acid required to react with the calculated moles of Mg(OH)₂:
moles of H₃PO₄ = (3/2) × moles of Mg(OH)₂
moles of H₃PO₄ = (3/2) × 0.003435 moles
moles of H₃PO₄ = 0.0051525 moles
Finally, we can calculate the volume of 1.07M phosphoric acid solution using the moles of H₃PO₄ calculated above:
volume (in L) = moles / concentration (in M)
volume (in L) = 0.0051525 moles / 1.07 M
volume (in L) = 0.004818 L
Therefore, the volume of 1.07M phosphoric acid solution necessary to react with 22.9 ml of 0.150M Mg(OH)₂ is approximately 0.004818 liters (or 4.818 ml).