For the reaction: 2 H3PO4 + 3 Ca(OH)2 → Ca3(PO4)2 + 6 H2O If you have 59.8 mL of 1.60 M phosphoric acid, how many mL of 0.500 M calcium hydroxide do you need to neutralize the acid?

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

Explanation:

Hello,

In this case, given the reaction:

We first compute the moles of phosphoric acid that are actually reacting given the volume (must be in litres) and the molarity:

Next, we compute the moles of calcium hydroxide that are reacting by using the 3:2 molar ratio with phosphoric acid:

Finally, by knowing the 0.500 M of the calcium hydroxide solution we compute the required volume:

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