As_{2}S_{3} a lyophobic colloid occurs as an orpiment of deep yellow red in color. H_{2}SO_{4} king of all acids is known as the oil of vitriol. Let us see how the As_{2}S_{3} reacts with the H_{2}SO_{4}.

**As _{2}S_{3} is a mono clinic crystal and a negatively charged sol with 3.46 g/cm^{3} density. It is used in glass, oil cloth, linoleum and intrinsic p – type semiconductors. Sulfur in H_{2}SO_{4} is sp^{3} hybridized with a C_{2 }point symmetry group and tetrahedral shape. It is reactive to metals, ionizes 100 %, and acts as a battery acid. **

In this article, we will discuss important facts about H_{2}SO_{4} + As_{2}S_{3} chemical reactions such as reaction enthalpy, the heat required, the product formed, the type of reaction, the type of intermolecular forces between their compounds, etc.

## What is the product of H_{2}SO_{4} and As_{2}S_{3}

**Arsenic acid (H _{3}AsO_{4}), Sulfur dioxide (SO_{2}), and H_{2}O (water)**

**are formed**

**when Arsenic trisulfide**

**(As**

_{2}S_{3})**reacts with sulfuric acid (H**.

_{2}SO_{4})**As _{2}S_{3} + 11 H_{2}SO_{4} → 2 H_{3}AsO_{4} + 14 SO_{2} + 8 H_{2}O**

## What type of reaction is H_{2}SO_{4} + As_{2}S_{3}

**H _{2}SO_{4} + As_{2}S_{3} falls under Double Displacement (Salt metathesis), redox, and exothermic reaction. **

## How to balance H_{2}SO_{4} + As_{2}S_{3}

**The unbalanced molecular equation for H_{2}SO_{4} + As_{2}S_{3} is.**

**As _{2}S_{3} + H_{2}SO_{4} →H_{3}AsO_{4} + SO_{2} +H_{2}O**

**To balance this equation, we should follow the steps given below:**

**Here, the number of H, O, As and S atoms is not the same on both sides of the reaction. So we will multiply these atoms with some coefficients so that they become equal.****The total number of As atoms on the reactant side is 2 while on the product side it is 1.****So, we multiply the**with a coefficient of 2 on the product side so that the number of As atoms become 2 on both sides of the reaction.**H**_{3}AsO_{4}**As**_{2}S_{3}+ H_{2}SO_{4}→ 2 H_{3}AsO_{4}+ SO_{2}+H_{2}O**Now,****2 H atoms are present on the reactant side while it is 8 on the product side of the reaction.****So, we multiply****H**with a coefficient of 11 on the reactant side and_{2}SO_{4}**H**multiply with 8 on the product side so that the number of H atoms becomes 22 on both sides of the reaction._{2}O**As**_{2}S_{3}+ 11 H_{2}SO_{4}→ 2 H_{3}AsO_{4}+ SO_{2}+8 H_{2}O**14 S and 44 O atoms are present on the reactant side while it is 1 on the product side so we multiply SO**_{2}with a 14 coefficient so that the number of S and O atoms becomes 14 and 44 respectively on both sides of the reaction.**Finally, the balanced equation is**:**As**_{2}S_{3}+ 11 H_{2}SO_{4}→ 2 H_{3}AsO_{4}+ 14 SO_{2}+ 8 H_{2}O

## H_{2}SO_{4} + As_{2}S_{3} titration

**The titration of H _{2}SO_{4} is not valid against As_{2}S_{3}. **

## H_{2}SO_{4} + As_{2}S_{3} net ionic equation

**The net ionic equation of H _{2}SO_{4} + As_{2}S_{3} is as follows**:

** As_{2}S_{3} (s) +16 H^{+} (aq) +11SO_{4}^{2–} (aq) = 6 AsO_{4}^{−} (aq)** +

**14SO**_{2}(g)**+**

**H**

_{2}O (l)**To get the net ionic equation for H_{2}SO_{4} + As_{2}S_{3}**,

**we should follow the steps given below:**

**Write the general balanced molecular equation.****As**_{2}S_{3}+ 11 H_{2}SO_{4}→ 2 H_{3}AsO_{4}+ 14 SO_{2}+ 8 H_{2}O**Now****the solubility equation for**is written**H**_{2}SO_{4}+ As_{2}S_{3}**by labeling the state or phase (s, l, g or aq) of each substance in the balanced molecular equation of**.**H**_{2}SO_{4}+ As_{2}S_{3}**As**_{2}S_{3}(s) + 11 H_{2}SO_{4}(aq) → 2 H_{3}AsO_{4}(aq) + 14 SO_{2}(g) + 8 H_{2}O**(l)****Break all aquatic soluble ionic substances into their corresponding ions to get the balanced ionic equation**.++22 H**As**_{2}S_{3}(s)^{+}(aq) +11SO_{4}^{2–}(aq) = 6H^{+}(aq)+6**AsO**_{4}^{−}(aq)**14SO**_{2}(g)**+****H**_{2}O(l)**To get the net ionic equation, remove spectator ions (H**) from the reactant and product side of the balanced ionic equation.^{+}**Finally, the net ionic equation for**:**H**is_{2}SO_{4}+ As_{2}S_{3}++16 H**As**_{2}S_{3}(s)^{+}(aq) +11SO_{4}^{2–}(aq) = 6**AsO**_{4}^{−}(aq)**14SO**_{2}(g)**+****H**_{2}O (l)

## H_{2}SO_{4} + As_{2}S_{3} conjugate pairs

**The ****conjugate pairs**** (compounds differ by one proton in their respective pair) in H_{2}SO_{4} + As_{2}S_{3} are:**

**The conjugate base of**acid is HSO**H**_{2}SO_{4}_{4}^{–}.**The conjugate base of H**_{2}O is OH^{–}.**The conjugate base of****H**_{3}AsO_{4}**is****H**_{2}AsO_{4}^{–}.**As**_{2}S_{3}**and**do not have their conjugate pairs**SO**_{2}**because both compounds do not contain hydrogen atom that can remove as proton ion.**

## H_{2}SO_{4} and As_{2}S_{3} intermolecular forces

**The ****intermolecular forces**** that work on ****H _{2}SO_{4}, As_{2}S_{3}**

**, H**

_{3}

**AsO**

_{4},

**SO**

_{2}**and**

**H**

_{2}O are-**Dipole-dipole force, London dispersion force, and hydrogen bonding are present in**molecules**H**_{2}SO_{4}**due to their polar and unsymmetrical nature.****Van der Waals forces are present in polar covalent As**._{2}S_{3}**The hydrogen bonding****present in H**due to three OH groups present in it._{3}AsO_{4}is**The dipole dipole interaction is present in SO**_{2}as it is a polar molecule.**Hydrogen bonds, dipole-induced dipole forces, and London dispersion forces****exist in H**_{2}O due to their strong polar and ionic nature.

## H_{2}SO_{4} + As_{2}S_{3} reaction enthalpy

**The net enthalpy change of the reaction H _{2}SO_{4} + As_{2}S_{3} is -28.60 kJ/mol. The value is gained from the following mathematical calculation.**

Compound | Standard Formation Enthalpy (Δ_{f}H°(Kj/mol)) |
---|---|

H_{2}SO_{4} | -735.13 |

As_{2}S_{3} | -169 |

H_{3}AsO_{4} | -904.6 |

H_{2}O | -285.83 |

SO_{2} | -297 |

**Standard Formation Enthalpy of Compounds**

**ΔH°**_{f}= ΣΔH°_{f}(products) – ΣΔH°_{f}(reactants)**(kJ/mol)****ΔH**_{f}= [2*( -904.6 ) +8 *(-285.83) +14 *(-297) -(11* (-735.13) +(-169 )] kJ/mol**ΔH**_{f}= -28.60 kJ/mol

## Is H_{2}SO_{4} + As_{2}S_{3} a buffer solution

**As _{2}S_{3} + H_{2}SO_{4} is not a buffer solution because H_{2}SO_{4}**

**is a strong acid and a buffer solution contains a weak acid and its conjugate base or a weak base and its conjugate acid.**

## Is H_{2}SO_{4} + As_{2}S_{3} a complete reaction

**As _{2}S_{3} + H_{2}SO_{4}**

**is a complete reaction because in this reaction**

**As**_{2}S_{3}and H_{2}SO_{4}**are completely consumed and converted into the final product (**

**H**successfully._{3}AsO_{4}, SO_{2})## Is H_{2}SO_{4} + As_{2}S_{3} an exothermic or endothermic reaction

** As_{2}S_{3} + H_{2}SO_{4}** reaction is an exothermic reaction because the net change of enthalpy is negative

**(i.e., ΔH**kJ/mol)

_{f}< 0, -28.60**where the -ve sign interprets the following facts about the reaction:**

**28.60 kJ/mol****heat is****released**by the**reactants**due to the formation of less energetic acid**As**_{2}S_{3}and H_{2}SO_{4}.**H**_{3}AsO_{4}-
**Heat emission by**rises**As**_{2}S_{3}and H_{2}SO_{4}**the energy of surroundings and makes the products stable**.

## Is H_{2}SO_{4} + As_{2}S_{3} a redox reaction

**H _{2}SO_{4} + As_{2}S_{3} is a redox reaction where**

**As**

_{2}S_{3}is a reducing agent, H_{2}SO_{4}is an oxidizing agent,**and the electron transportation**

**occurs as follows:**

**As**^{+3 }**– 2 e**^{–}→ As^{+5}(oxidation)**S**^{-2}– 6 e^{–}→ S^{+4}(oxidation)**S**^{+6}+ 2 e^{–}→ S^{+4}(reduction)

## Is H_{2}SO_{4} + As_{2}S_{3} a prcipitation reaction

**As_{2}S_{3} + H_{2}SO_{4}** is not a precipitation reaction

**because the completion of the reaction provides the**

**H**_{3}AsO_{4}**as the main product that is dissolved in reaction media.**

## Is H_{2}SO_{4} + As_{2}S_{3} reversible or irreversible reaction

**As_{2}S_{3} + H_{2}SO_{4}** is an irreversible reaction because the products

**are stable so they do not need to react with each other to form reactants back.**

**H**_{3}AsO_{4}and SO_{2}## Is H_{2}SO_{4} + As_{2}S_{3} displacement reaction

**As_{2}S_{3} + H_{2}SO_{4} **is a double displacement reaction because, in this reaction, sulfate ion (SO

_{4}

^{2–}) and sulfide ion (S

^{2-}) exchange their places with each other to form new products,

**.**

**H**_{3}AsO_{4}and SO_{2}#### Conclusion

This article concludes that As_{2}S_{3} + H_{2}SO_{4} reaction is carried out through a double displacement mechanism and produces more stable H_{3}AsO_{4} and SO_{2} by emitting approximately 28 KJ/mol energy. The As_{2}S_{3} + H_{2}SO_{4 }reaction displays the cyclic voltammograms due to electron transportation.