Let us learn about a chemical reaction with unique mechanisms, exciting chemistry, and properties. Below is an explanation of the reaction between sulfuric acid and sodium sulphide.

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**Sulphuric acid (H _{2}SO_{4}) is also known as the king of acids due lớn its extensive use in producing many compounds. Sodium sulphide, or Na_{2}S, is a well-known inorganic chemical compound with the formula Na_{2}S and acts as a base.**

Now we can discuss the mechanism of the reaction, such as enthalpy, type of reaction, intermolecular force, conjugate pairs, product formation, etc., in the subsequent article.

**What is the product of H**_{2}SO_{4} and Na_{2}S

_{2}SO

_{4}and Na

_{2}S

**H _{2}SO_{4} and Na_{2}S**

**will be produced by the reaction between phosphoric acid and sodium sulphide as mentioned in the below-mentioned reaction.**

**H**_{2}SO_{4 }(aq)+Na_{2}S (s) = Na_{2}SO_{4 }(s)+ H_{2}S (aq)

**What type of reaction is H**_{2}SO_{4} and Na_{2}S

_{2}SO

_{4}and Na

_{2}S

**H _{2}SO_{4} and Na_{2}S are a double displacement reaction**

**when the cation (sodium) of sodium sulphide and the anion (Sulfate) of sulphuric acid switch positions, and the formation of new compounds takes place.**

**How lớn balance H**_{2}SO_{4} and Na_{2}S

_{2}SO

_{4}and Na

_{2}S

**H _{2}SO_{4} and Na_{2}S balance reactions are as follows**:

**H _{2}SO_{4 }+ Na_{2}S = Na_{2}SO_{4 }+ H_{2}S**

**Further,** **the reaction between H _{2}SO_{4} and Na_{2}S is balanced using the following steps:**

**First, the reaction between the aforementioned reactants must be established.**

**H**_{2}SO_{4 }+ Na_{2}S = Na_{2}SO_{4 }+ H_{2}S

**Coefficient of the individual chemical molecules are labeled as,**

**H**_{2}SO_{4}(a), Na_{2}S (b), Na_{2}SO_{4 }(c)and H_{2}S (d)

**Count the number of hydrogen (H), Sulphur (S), oxygen (O), and sodium (Na) on both the reactant and product sides. That gives**

·

Coefficient at Reactant side | Element | Coefficient at the Product side |

2 | S | 2 |

4 | O | 4 |

2 | H | 2 |

2 | Na | 2 |

**Coefficient of reaction on both sides of the reaction**

**Now equate the number of elements****from both the reactant and product sides.****This yields the balance equation****H**_{2}SO_{4 }+ Na_{2}S = Na_{2}SO_{4 }+ H_{2}S

**H**_{2}SO_{4} + Na_{2}S titration

_{2}SO

_{4}+ Na

_{2}S titration

**H _{2}SO_{4 }and Na_{2}S will yield acid base titration when a standard solution of Na_{2}S is titrated against H_{2}SO_{4} in the presence of an indicator. Several apparatus, conditions, and procedures for the titration are discussed below:**

**Apparatus used**

**Burette, burette stand, volumetric flask, conical flask, and beakers.**

**Indicator**

**Bromophenol xanh lơ serves as an indicator**

**Procedure**

**First, wash all the apparatus with clean water, and dry it in the oven****Then make known the normal solution of Na**S and H_{2}SO_{2}_{4}using molar equations**Fill the burette with acid till the 50ml mark and take 10 ml of Na**_{2}S in a conical flask.**Add a few (3) drops of indicator in a conical flask.****Titrate the base against acid by slowly adding acid into it dropwise.****Mark the reading of the burette when colure of the solution turns yellow lớn xanh lơ.****The reading will tell about the neutralization point of the solution.****Repeat the process in triplicate lớn record concordant readings.**

**H**_{2}**SO**_{4}** + Na**_{2}**S net ionic equation**

_{2}

_{4}

_{2}

**H _{2}SO_{4} and Na_{2}S net ionic equation of is**

**2H ^{+} + S^{–} = H_{2}S^{+}**

**Separate reactants and products into ions lớn determine which are spectators. Which is****Na**_{2}S (aq) + H_{2}SO_{4}(aq) = Na_{2}SO_{4}(aq) + H_{2}S (g)**Here first split the strong electrolytes into ions (the complete ionic equation) which is****2H**^{+}+ SO_{4}^{2-}+ 2Na^{+}+ S^{–}= H_{2}S + 2Na^{+}+ SO_{4}^{2-}**Now cross out the spectator ions on both sides of the ionic equation****2H**^{+}+ S^{–}= H_{2}S^{+}**This yield the net ionic equation, which is****2H**^{+}+ S^{–}= H_{2}S^{+}

**H**_{2}SO_{4} + Na_{2}S conjugate pairs

_{2}SO

_{4}+ Na

_{2}S conjugate pairs

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**H _{2}SO_{4} and Na_{2}S reaction has the following conjugate pair.**

**The conjugate base of H**_{2}SO_{3}is HSO^{3–}(bisulfite anion)**Conjugate pair of base Na**_{2}S is Na^{2+}(sodium ion)

**H**_{2}SO_{4} and Na_{2}S intermolecular forces

_{2}SO

_{4}and Na

_{2}S intermolecular forces

**H _{2}SO_{4} and Na_{2}S reaction has the following intermolecular forces,**

**H**_{2}SO_{4}and Na_{2}S**element,****H**_{2}SO_{4}is bonded via cầu xin der Waals dispersion forces, dipole-dipole interactions, and hydrogen bonding.**Na**_{2}S has positive lớn negative intermolecular forces due lớn its ionic bond.

**H**_{2}**SO**_{4}** and Na**_{2}**S reaction enthalpy**

_{2}

_{4}

_{2}

**H _{2}SO_{4} and Na_{2}S reaction enthalpy is -296.83 KJ/mol which is negative as the energy has been released from the reaction.**

**Is H**_{2}**SO**_{4}** + Na**_{2}**S a buffer solution**

_{2}

_{4}

_{2}

**H _{2}SO_{4} and Na_{2}S can not make a buffer, as H_{2}SO_{4 }is a strong acid and Na_{2}S is a strong base in neutral conditions.**

**Is H**_{2}**SO**_{4}** + Na**_{2}**S a complete reaction**

_{2}

_{4}

_{2}

**H _{2}SO_{4} and Na_{2}S are a complete reaction as they utilize entirely one of the reactants.**

**Is H**_{2}**SO**_{4}** + Na**_{2}**S an exothermic or endothermic reaction**

_{2}

_{4}

_{2}

**H _{2}SO_{4} and Na_{2}S** reaction is exothermic in nature.

**The reaction of both elements will release heat in the khuông of energy and can be calculated.**

Element | Energy in kJ/mol for element |

Na_{2}S | -369 |

H_{2}SO_{4} | -814 |

Na_{2}SO_{4} | -287.8 |

H_{2}S | -20 |

**The energy of individual molecules for the calculation of enthalpy of the reaction**

**Reactant = -369 kJ/mol + 814 kJ/mol = 1183****Product = 287.8 kJ/mol + -20 kJ (g) = 307.8****Enthalpy= Product – reactant = -875.2****Which is negative, resulting in the release of energy resulting in an exothermic reaction.**

**Is H**_{2}SO_{4} and Na_{2}S a redox reaction

_{2}SO

_{4}and Na

_{2}S a redox reaction

**H _{2}SO_{4} and Na_{2}S is not redox when phosphoric acid and hydrochloric acid react. **

**+2 -1 +2 -1 +2-1 +2-1**

**Na _{2}S + H_{2}SO_{4} = Na_{2}SO_{4} + H_{2}S **

**Is H**_{2}SO_{4} + Na_{2}S a precipitation reaction

_{2}SO

_{4}+ Na

_{2}S a precipitation reaction

**H _{2}SO_{4} and Na_{2}S reactions are not precipitation reactions as**

**h**

**ydrogen sulphide gas that has been produced is being released.**

**Is H**_{2}SO_{4} + Na_{2}S reversible or irreversible reaction

_{2}SO

_{4}+ Na

_{2}S reversible or irreversible reaction

**H _{2}SO_{4} + Na_{2}S reaction is not a reversible reaction. The equilibrium cannot be shifted backward as H_{2}S is released from the product.**

**Is H**_{2}SO_{4} + Na_{2}S displacement reaction

_{2}SO

_{4}+ Na

_{2}S displacement reaction

**H _{2}SO_{4} and Na_{2}S reaction is a displacement reaction because of another displaced element from each compound. **

**Conclusion**

The production of sodium sulphate and the evolution of hydrogen sulphide gas resulted from the displacement reaction between sodium sulphide, a salt, and sulphuric acid, a strong acid. Exothermic in nature, the process releases energy and is irreversible in nature.

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