Answer:
V = 127.57 mL
Explanation:
In this case, we are having an acid base titration, and this, according to the overall reaction:
2H₃PO₄ + 3Ca(OH)₂ -----> Ca₃(PO₄)₂ + 6H₂O
This reaction was neutralized, so, it's a neutralization reaction, therefore, we can use the following expression for neutralization reactions:
n₁ = n₂ (1)
This is because, in the equivalence point, both moles of the reagents are the same. Now we need to do a relation between this and the actual moles that we have according to the overall reaction.
In this case, we have 2 moles of acid reacting with 3 moles of base, so, the equation (1) is re-written like this:
2n₁ = 3n₂ (2)
Knowing this, and that we have values of concentrations and volume, we can write the expression of moles in terms of molarity and volume:
n = M * V replacing in (2):
2M₁V₁ = 3M₂V₂ From here, we can solve for V₂ which is the volume of the base:
V₂ = 2M₁V₁ / 3M₂ (3)
Now, all we have to do is replace the given values to obtain the volume of the base:
V₂ = 2 * 1.6 * 59.8 / 3 * 0.5
<h2>
V₂ = 127.57 mL</h2>
