Question

Question 3.) Calculate the grams of phosphorus in the 100 ml sample

The acid-base titration procedure depends on the ability of a certain stoichiometric amount of sodium hydroxide being capable of transforming the precipitate as follows:

X OH^-   +    (NH₄ )₃[P(Mo₁₂O₄₀)]×12H₂O -> 3 NH₄⁺ + HP0₄^2- + 12 MoO₄^2-   +   Y H₂O

Balance this equation to find numerical values for X and Y.

50.00 mL of 0.1000 M sodium hydroxide is added to the sample, enough to have enough NaOH to react plus some excess. Titration of the excess NaOH was carried out with standard HCl and was executed to the point where the color of phenolphthalein just dis­appeared. After adding the NaOH, it required 9.06 mL of 0.1000 M HC1 to reach the endpoint.

Answer 1

Start by writing out the equation as given.

X OH^- + (NH₄)₃[P(Mo₁₂O₄₀)].12H₂O --------> 3 NH₄⁺ + HPO₄^2- + 12 MoO₄^2- + Y H₂O

We have 36 H (except OH^-) on the left and 13 on the right (except H₂O). There is a deficit of 23 H on the right; to balance, add 23 H₂O on the right. This gives

X OH^- + (NH₄)₃[P(Mo₁₂O₄₀)].12H₂O --------> 3 NH₄⁺ + HPO₄^2- + 12 MoO₄^2- + 23 H₂O

The number of H on the right is 59 while that on the left is 36 (except OH^-). Add 23 OH^- on the left to get the balanced equation as

23 OH^- + (NH₄)₃[P(Mo₁₂O₄₀)].12H₂O --------> 3 NH₄⁺ + HPO₄^2- + 12 MoO₄^2- + 23 H₂O

The above equation is balanced (ans).

As per the stoichiometric equation above,

23 mole OH^- = 1 mole HPO₄^-

Millimoles of NaOH added = (50.00 mL)*(0.1000 M) = 5.000 mmole.

NaOH reacts with HCl as per the equation

NaOH (aq) + HCl (aq) -------> NaCl (aq) + H₂O (l)

As per the stoichiometric equation,

1 mole NaOH = 1 mole HCl.

Millimoles of HCl added = millimoles of excess NaOH = (9.06 mL)*(0.1000 M) = 0.906 mmole.

Millimoles of NaOH that gives HPO₄^- = (5.000 – 0.906) mmole = 4.094 mmole.

Millimoles of HPO₄^- formed = (4.094 mmole NaOH)*(1 mole HPO₄^-/23 mole NaOH) = 0.178 mmole.

Now 1 mole HPO₄^- = 1 mole P.

Atomic mass of P = 30.974 g/mol.

Mass of P in the sample = (0.178 mmole)*(1 mole/1000 mmole)*(30.974 g/mol) = 0.005513372 g ≈ 0.005513 g (ans).