Sulfuric acid is a mineral acid, highly viscous as absorbs moisture from air, and Sodium bicarbonate is a crystalline powder. Let us discuss some reactions of H_{2}SO_{4 }and NaHCO_{3}.

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**H _{2}SO_{4} is a strong colorless acid with high viscosity, miscible in water with another name oil of vitriol, used for mineral processing, chemical analysis, and oil refining, and NaHCO_{3} is a white crystalline salt which is highly soluble in water and used if food industries as a baking powder.**

In the following parts of this article, we will talk about the reaction enthalpy of H_{2}SO_{4} + NaHCO_{3} with conjugate pairs, net ionic equation, conjugate pairs, type of reaction, etc.

**What is the product of H**_{2}SO_{4} and NaHCO_{3}

_{2}SO

_{4}and NaHCO

**The product of H _{2}SO_{4} + NaHCO_{3} is**

**Na**

_{2}SO_{4 }**with the liberation of carbon dioxide CO**

_{2}and water H_{2}O.

**H _{2}SO_{4} + NaHCO**

_{3}

**=**

**Na**+ CO_{2}SO_{4 }_{2}+ H_{2}O.

**What type of reaction is H**_{2}SO_{4} + NaHCO_{3}

_{2}SO

_{4}+ NaHCO

**H _{2}SO_{4} + NaHCO**

_{3}is a double displacement reaction.

**How to tướng balance ****H**_{2}SO_{4} + NaHCO_{3}

**H**

_{2}SO_{4}+ NaHCO_{3}

**Here, are the steps to tướng balance the reaction of H_{2}SO_{4} + NaHCO_{3}**

** H_{2}SO_{4} + 2 NaHCO_{3}** =

**Na**+2 CO_{2}SO_{4 }_{2}+ 2 H_{2}O

**Name the reactant and products with alphabets A, B, C, D, and E**

**A**= C**H**_{2}SO_{4}+ B NaHCO_{3}**Na**+D CO_{2}SO_{4 }_{2}+ E H_{2}O.

**Modify the atoms with a suitable number**

**H –>A, B, E, S –>A, C, D, O –> A, B, C, D, E Na –>B, C, C –> B, D**

**Multiply the coefficients with a suitable number**

**A = 1, B = 2, C = 1, D =2, E = 2**

**To write the balanced equation reduce the lowest integer value.**

=**H**_{2}SO_{4}+ 2 NaHCO_{3}**Na**+2 CO_{2}SO_{4 }_{2}+ 2 H_{2}O.

**H**_{2}SO_{4} + NaHCO_{3} titration

**titration**

**H**_{2}SO_{4}+ NaHCO_{3}** H_{2}SO_{4}** cannot be titrated with

**because during the reaction CO**

**NaHCO**_{3}_{2}gas is librated with water and it is not possible to tướng calculate the endpoint and the unknown concentration of the NaHCO

_{3}.

**H**_{2}SO_{4} + NaHCO_{3} net ionic equation

**net ionic equation**

**H**_{2}SO_{4}+ NaHCO_{3}**The net ionic equation of the reaction H_{2}SO_{4} + NaHCO_{3} is –**

**2H ^{+} + SO^{4-} + Na^{+} + HCO^{3-} = Na^{+} + SO^{4-} + CO_{2} + H^{+} + OH^{– } **

**Write the full reaction with the respective states of molecules**

**(l)**= (s)**H**_{2}SO_{4}+ (s) NaHCO_{3}**Na**+(g) CO_{2}SO_{4 }_{2}+ (l) H_{2}O.

**Split the atoms into ions thus, the net ionic equation is –**

**2H**^{+}+ SO^{4-}+ Na^{+}+ HCO^{3-}= Na^{+}+ SO^{4-}+ CO_{2}+ H^{+}+ OH^{– }

**H**_{2}SO_{4} + NaHCO_{3} conjugate pairs

**conjugate pairs**

**H**_{2}SO_{4}+ NaHCO_{3}**H _{2}SO_{4} + NaHCO_{3} has the following conjugate pairs, **

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**The conjugate base pair is HSO**^{4-}after the protonation of H_{2}SO_{4}.

**The conjugate acid pair is H**_{2}CO_{3}after the deprotonation of NaHCO_{3}.

**H**_{2}SO_{4} and NaHCO_{3} intermolecular forces

**intermolecular forces**

**H**_{2}SO_{4}and NaHCO_{3}**H _{2}SO_{4} + NaHCO_{3} has the following intermolecular forces,**

**The strong electrostatic force with covalent bonds is the intermolecular force present in H**_{2}SO_{4.}**A covalent bond and coordinate bonds with intermolecular hydrogen bonds are the intermolecular forces present in NaHCO**_{3}.

**H**_{2}SO_{4} + NaHCO_{3} reaction enthalpy

**reaction enthalpy**

**H**_{2}SO_{4}+ NaHCO_{3}**The reaction enthalpy of H_{2}SO_{4} + NaHCO_{3} is -133 KJ/mol.**

Molecule | Enthalpy |

H_{2}SO_{4} | -814 KJ/mol |

NaHCO_{3} | -947 KJ/mol |

**Table showing enthalpy of formation**

**The total enthalpy is**–

- (
**-814 KJ/mol) – (-947 KJ/mol) = -133 KJ/mol.**

**Is ****H**_{2}SO_{4} + NaHCO_{3} a buffer solution

**a buffer solution**

**H**_{2}SO_{4}+ NaHCO_{3}**H_{2}SO_{4} + NaHCO_{3}** is not a buffer solution because NaHCO

_{3}nature is amphoteric and it will not allow the pH to tướng increase more than vãn 7.

**Is ****H**_{2}SO_{4} + NaHCO_{3} a complete reaction

**a complete reaction**

**H**_{2}SO_{4}+ NaHCO_{3}** H_{2}SO_{4} + NaHCO_{3} is a complete reaction. The product formed is Na_{2}SO_{4} which is a whole chemical compound and complex and cannot be reacted further after the liberation of CO_{2} gas.**

**Is ****H**_{2}SO_{4} + NaHCO_{3} an exothermic or endothermic reaction

**an exothermic or endothermic reaction**

**H**_{2}SO_{4}+ NaHCO_{3}**H_{2}SO_{4} + NaHCO_{3}** is an exothermic reaction as the product is

**Na**with the formation of bubbles of CO

_{2}SO_{4}_{2}gas which increases the temperature of the solution and gets heated up.

**Is ****H**_{2}SO_{4} + NaHCO_{3} a redox reaction

**a redox reaction**

**H**_{2}SO_{4}+ NaHCO_{3}**H_{2}SO_{4} + NaHCO_{3}** is not a redox reaction as all the atoms of reactants and products are in the same oxidation states.

**Is ****H**_{2}SO_{4} + NaHCO_{3} a precipitation reaction

**a precipitation reaction**

**H**_{2}SO_{4}+ NaHCO_{3}** H_{2}SO_{4} + NaHCO_{3} is not a precipitation reaction as the product is Na_{2}SO_{4} which is highly soluble in water and cannot be precipitated down in the solution.**

**Is **** ****H**_{2}SO_{4} + NaHCO_{3} reversible or irreversible reaction

**reversible or irreversible reaction**

**H**_{2}SO_{4}+ NaHCO_{3}** H_{2}SO_{4} + NaHCO_{3}** is not reversible as the product is

**and it is highly soluble in water and cannot be reversed as a reactant.**

**Na**_{2}SO_{4}

**Is**** ****H**_{2}SO_{4} + NaHCO_{3} displacement reaction

**displacement reaction**

**H**_{2}SO_{4}+ NaHCO_{3}** H_{2}SO_{4} + NaHCO_{3}** is a double displacement reaction.

**2H**

^{+}ion displaces the Na to tướng form**Na**and water with CO_{2}SO_{4 }_{2}gas as, two different compounds are formed.

**Conclusion**

H_{2}SO_{4} is a strong acid that can be used for the refining of sugar and oil and is highly reactive toward acids and bases of chemical compounds and can analyze their properties too on the other side NaHCO_{3} is used as baking powder in the food industry.

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