Dissolving 1.24 g of white phosphorous in boiling NaOH solution in an inert atmosphere gives a gas Q

Q: Why does white phosphorus react with NaOH but not red phosphorus?

White phosphorus consists of discrete $P_4$ tetrahedra with high angular strain, making it highly reactive. Red phosphorus is polymeric and much more stable, so it doesn't react with boiling NaOH.

Q: How do I remember the products of $P_4$ and NaOH?

Remember it is a disproportionation reaction. Phosphorus (oxidation state $0$) goes to $-3$ in Phosphine ($PH_3$) and $+1$ in Sodium hypophosphite ($NaH_2PO_2$).

Q: What is the color of the $Cu_3P_2$ precipitate?

Copper(II) phosphide ($Cu_3P_2$) forms as a characteristic black precipitate when phosphine gas is passed through a copper sulfate solution.

Q: Can I use the equivalent concept instead of moles here?

Yes, you can use equivalents, but since the n-factors change across different reactions (disproportionation vs precipitation), using the mole method with balanced equations is much safer and less prone to errors.

Q: Why is an inert atmosphere required for the first reaction?

White phosphorus is highly reactive and can spontaneously catch fire in the presence of oxygen to form $P_4O_{10}$. An inert atmosphere prevents this unwanted oxidation.

Question

Dissolving 1.24 g of white phosphorous in boiling NaOH solution in an inert atmosphere gives a gas Q. The amount of CuSO4 (in g) required to completely consume the gas Q is ______.
[Given : Atomic mass of H = 1, O = 16, Na = 23, P = 31, S = 32, Cu = 63]

Sigma Percentile · Accepted Answer

Imagine you are standing in a chemistry lab, holding a flask containing white phosphorus. White phosphorus, with its highly strained $P_4$ tetrahedral structure, is notoriously reactive. When we drop it into boiling sodium hydroxide ($NaOH$) in an inert atmosphere, a fascinating chemical dance begins.

### The Master Equation: Disproportionation

This is a classic **disproportionation reaction**. Phosphorus, initially at an oxidation state of $0$, simultaneously oxidizes and reduces itself. It forms phosphine gas ($PH_3$) where its oxidation state drops to $-3$, and sodium hypophosphite ($NaH_2PO_2$) where it rises to $+1$.

The balanced equation is:
$$P_4 + 3NaOH + 3H_2O ightarrow PH_3 + 3NaH_2PO_2$$

From this, we can clearly see that $1$ mole of $P_4$ yields exactly $1$ mole of $PH_3$.

Now, let's calculate the moles of $P_4$ we started with. The molar mass of $P_4$ is $4 	imes 31 = 124 	ext{ g/mol}$.

$$n_{P_4} = \frac{1.24}{124} = 0.01 	ext{ mol}$$

Therefore, the moles of gas Q ($PH_3$) produced is also $0.01 	ext{ mol}$.

### The Master Equation: Precipitation

Next, we pass this phosphine gas into a solution of copper sulfate ($CuSO_4$). Phosphine is a weak base and a reducing agent, but here it reacts to form a black precipitate of copper(II) phosphide ($Cu_3P_2$).

The balanced equation for this precipitation is:
$$2PH_3 + 3CuSO_4 ightarrow Cu_3P_2 \downarrow + 3H_2SO_4$$

Notice the stoichiometry here. It takes $3$ moles of $CuSO_4$ to completely consume $2$ moles of $PH_3$.

$$n_{CuSO_4} = \frac{3}{2} 	imes n_{PH_3}$$

Substituting the moles of $PH_3$ we found earlier:

$$n_{CuSO_4} = \frac{3}{2} 	imes 0.01 = 0.015 	ext{ mol}$$

### Final Calculation

Finally, we need to find the mass of $CuSO_4$ required. First, we calculate its molar mass.

$$M_{CuSO_4} = 63 + 32 + (4 	imes 16) = 159 	ext{ g/mol}$$

Now, we simply multiply the moles by the molar mass to get our final answer.

$$W_{CuSO_4} = 0.015 	imes 159 = 2.385 	ext{ g}$$

And there we have it! A beautiful sequence of reactions perfectly quantified by stoichiometry.

Enter Numerical Value:

Visualized Solution

Reaction Setup

Reaction 1: Disproportionation of $P_{4}$

Moles of $P_{4}$

Moles of Gas Q ( $P H_{3}$ )

Reaction 2: Phosphine with $C u S O_{4}$

Moles of $C u S O_{4}$ Required

Molar Mass of $C u S O_{4}$

Mass of $C u S O_{4}$

The Way Forward

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Master Equation

The Master Equation

Final Calculation

Similar Questions

The volume (in $mL$ ) of $0.1 N NaOH$ required to neutralise $10 mL$ of $0.1 N$ phosphonic acid is ............ .

To neutralise completely $20 mL$ of $0.1 M$ aqueous solution of phosphorous acid ( $H_{3} PO_{3}$ ), the volume of $0.1 M$ aqueous $KOH$ solution required is

The minimum amount of $O_{2} (g)$ consumed per gram of reactant is for the reaction (Given atomic mass : $Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1$ )

What volume of hydrogen gas, at $273 K$ and $1 atm$ pressure will be consumed in obtaining $21.6 g$ of elemental boron (atomic mass $= 10.8$ ) from the reduction of boron trichloride by hydrogen?

The ratio of the mass percentages of 'C' and 'H' and 'C' and 'O' of a saturated acyclic organic compound 'X' are $4 : 1$ and $3 : 4$ respectively. Then, the moles of oxygen gas required for complete combustion of two moles of organic compound 'X' is ......... .

The exact volumes of $1 M NaOH$ solution required to neutralise $50 mL$ of $1 M H_{3} PO_{3}$ solution and $100 mL$ of $2 M H_{3} PO_{2}$ solution, respectively, are

The mass of potassium dichromate crystals required to oxidise $750 cm^{3}$ of $0.6 M$ Mohr's salt solution is (molar mass $= 392$ )

Enter Numerical Value:

Visualized Solution

Reaction Setup

Reaction 1: Disproportionation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P4​

Moles of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P4​

Moles of Gas Q ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PH3​)

Reaction 2: Phosphine with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CuSO4​

Moles of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CuSO4​ Required

Molar Mass of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CuSO4​

Mass of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CuSO4​

The Way Forward

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Master Equation

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphonic acid is ............ .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3​PO3​), the volume of 0.1 M aqueous KOH solution required is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The minimum amount of O2​(g) consumed per gram of reactant is for the reaction (Given atomic mass : Fe=56,O=16,Mg=24,P=31,C=12,H=1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H3​PO3​ solution and 100 mL of 2 M H3​PO2​ solution, respectively, are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The mass of potassium dichromate crystals required to oxidise 750 cm3 of 0.6 M Mohr's salt solution is (molar mass =392)

Reaction 1: Disproportionation of $P_{4}$

Moles of $P_{4}$

Moles of Gas Q ( $P H_{3}$ )

Reaction 2: Phosphine with $C u S O_{4}$

Moles of $C u S O_{4}$ Required

Molar Mass of $C u S O_{4}$

Mass of $C u S O_{4}$

The volume (in $mL$ ) of $0.1 N NaOH$ required to neutralise $10 mL$ of $0.1 N$ phosphonic acid is ............ .

To neutralise completely $20 mL$ of $0.1 M$ aqueous solution of phosphorous acid ( $H_{3} PO_{3}$ ), the volume of $0.1 M$ aqueous $KOH$ solution required is

The minimum amount of $O_{2} (g)$ consumed per gram of reactant is for the reaction (Given atomic mass : $Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1$ )

The exact volumes of $1 M NaOH$ solution required to neutralise $50 mL$ of $1 M H_{3} PO_{3}$ solution and $100 mL$ of $2 M H_{3} PO_{2}$ solution, respectively, are

The mass of potassium dichromate crystals required to oxidise $750 cm^{3}$ of $0.6 M$ Mohr's salt solution is (molar mass $= 392$ )