Question

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?

Answer 1

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

V₂ = 127.57 mL

Answer 2

Answer:

$V_{Ca(OH)_2}=0.287L=287mL$

Explanation:

Hello,

In this case, given the reaction:

$2 H_3PO_4 + 3 Ca(OH)_2 \rightarrow Ca_3(PO_4)_2 + 6 H_2O$

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

$n_{H_3PO_4 }=1.60\frac{mol}{L}*59.8mL*\frac{1L}{1000mL} =0.0957molH_3PO_4$

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

$n_{Ca(OH)_2}=0.0957molH_3PO_4*\frac{3molCa(OH)_2}{2molH_3PO_4} =0.144molCa(OH)_2$

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

$V_{Ca(OH)_2}=\frac{n_{Ca(OH)_2}}{M_{Ca(OH)_2}}=\frac{0.144mol}{0.500\frac{mol}{L} } \\ \\V_{Ca(OH)_2}=0.287L=287mL$

Best regards.