Sulphuric acid is a strong acid referred to lớn as the king of acids. In contrast, sodium nitrate is a neutral base. Let us study how H_{2}SO_{4} and NaNO_{3} react.

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**H _{2}SO_{4} + NaNO_{3} is**

**a salt metathesis reaction. Sulphuric acid is also known as the oil of vitriol. It is a transparent, odourless, viscous liquid. In contrast, sodium nitrate is a neutral salt formed by combining sodium hydroxide and nitric acid. It is a key ingredient of fertilisers.**

In this article we will study the type, product, and balancing of the H_{2}SO_{4} + NaNO_{3} reaction.

## What is the product of H_{2}SO_{4} and NaNO_{3}?

**Na _{2}SO_{4} (Sodium Sulphate) and HNO_{3} (Nitric Acid) are formed on the reaction of H_{2}SO_{4} and NaNO_{3}**.

**H**_{2}**SO**_{4}** + NaNO**_{3}** → Na**_{2}**SO**_{4}** + HNO**_{3}** **

## What type of Reaction is H_{2}SO_{4} + NaNO_{3}

**H _{2}SO_{4} + NaNO_{3} is a double displacement reaction where the ions of the ionic compounds exchange their site to lớn size two new compounds.**

## How to lớn Balance H_{2}SO_{4} + NaNO_{3}

**The equation is balanced using hit-and-trial method by using the following steps.**

**H _{2}SO_{4} + NaNO_{3} → Na_{2}SO_{4} + HNO_{3}**

**First, we will compare the reactants’ atoms with the product’s atoms**.

Reactant Side | Product Side |
---|---|

2 Hydrogen units from Sulphuric Acid | 1 Hydrogen atom unit Nitric Acid |

1 Sulphate unit from Sulphuric Acid | 1 Sulphate anion unit Sodium Sulphate |

1 Sodium unit from Sodium Nitrate | 2 Sodium atom units Sodium Sulphate |

1 Nitrate unit from Sodium Nitrate | 1 Nitrate unit from Nitric Acid |

**Comparison of the atoms of the reactant side with the product side**

**We can see that the sodium and hydrogen are imbalanced.****Thus, we will balance the sodium by putting coefficient 2 on the reactant side.****H**_{2}**SO**_{4}**+ 2NaNO**_{3}**→ Na**_{2}**SO**_{4}**+ HNO**_{3}**On putting the coefficient 2, the number of nitrate ions is also increased to lớn 2 on the reactant side.****Thus, we will balance it by putting in 2HNO**_{3}**.****H**_{2}**SO**_{4}**+ 2NaNO**_{3}**→ Na**_{2}**SO**_{4}**+ 2HNO**_{3}**On putting 2HNO**_{3}**, the number of hydrogen is also balanced.****Hence, we can summarise that the above equation is the required balanced equation.****H**_{2}**SO**_{4}**+ 2NaNO**_{3}**→ Na**_{2}**SO**_{4}**+ 2HNO**_{3}

## H_{2}SO_{4} + NaNO_{3} Titration

**H _{2}SO_{4} + NaNO_{3} does not undergo titration as there is no equivalence point.**

## H_{2}SO_{4} + NaNO_{3} Net Ionic Equation

**The net ionic equation of H _{2}SO_{4} + NaNO_{3} is **

**2H ^{+} + 2NO_{3}^{–} → 2HNO_{3}. **

**To derive the net ionic equation, we will follow the following steps.**

**Foremost, we will write the balanced equation to lớn calculate the net ionic equation.****H**_{2}SO_{4}(aq) + 2NaNO_{3}(s) → Na_{2}SO_{4}(aq) + 2HNO_{3}**(aq)****Hence, the total ionic equation for the above equation will be:****2H**^{+}+ SO_{4}^{2-}+ 2Na^{+}+ 2NO_{3}^{–}→ 2Na^{+}+ SO_{4}^{2-}+ 2HNO_{3}**2Na**^{+}and SO_{4}^{2-}are present in reactant and product, Thus, will get cancelled out.**Hence, the net ionic equation would be:****2H**^{+}+ 2NO_{3}^{–}→ 2HNO_{3}

## H_{2}SO_{4} + NaNO_{3} Conjugate Pairs

**The conjugate base of H**_{2}SO_{4}is sulphate ion (SO_{4}^{2-}).**NaNO**_{3}has no conjugate pair due to lớn the absence of protons.

## H_{2}SO_{4} and NaNO_{3} Intermolecular Forces

**H**_{2}SO_{4}contains hydrogen bonding, electrostatic force, nài der Waals force and dipole-dipole intermolecular force of attraction.**NaNO**_{3}contains electrostatic force and nài der Waals intermolecular force of attraction.

## Is H_{2}SO_{4} + NaNO_{3} a Buffer Solution

**H _{2}SO_{4} + NaNO_{3} is not a buffer solution, as it is a strong acid solution and a neutral salt.**

## Is H_{2}SO_{4} + NaNO_{3} a Complete Reaction

**H _{2}SO_{4} + NaNO_{3} is a complete reaction**

**as no reactant is left at the over of reaction.**

**H**_{2}**SO**_{4}** + 2NaNO**_{3}** → Na**_{2}**SO**_{4}** + 2HNO**_{3}

## Is H_{2}SO_{4} + NaNO_{3} an Endothermic Reaction

**H**_{2}**SO**_{4}** + NaNO**_{3}** is an endothermic reaction, as heat is supplied during the reaction and the value of delta G is more than vãn zero (positive).**

## Is H_{2}SO_{4} + NaNO_{3} a Redox Reaction

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**H _{2}SO_{4} + NaNO_{3} is not a redox reaction as the compound does not undergo oxidation or reduction.**

## Is H_{2}SO_{4} + NaNO_{3} a Precipitation Reaction

**H _{2}SO_{4} + NaNO_{3} is not a precipitation reaction, as no insoluble residue is formed.**

## Is H_{2}SO_{4} + NaNO_{3} Reversible Reaction

**H _{2}SO_{4} + NaNO_{3} is a reversible reaction as we can get products from the reactant.**

## Is H_{2}SO_{4} + NaNO_{3} Displacement Reaction

**H _{2}SO_{4} + NaNO_{3} is a displacement reaction as the sulphate ion displaces the nitrate ion from sodium nitrate to lớn size sodium sulphate. And the hydrogen ion replaces the sodium ion from sodium nitrate to lớn size nitric acid.**

**H _{2}SO_{4} + 2NaNO_{3} → Na_{2}SO_{4} + 2HNO_{3}**

## How to lớn Balance NaNO_{3} + Cu + H_{2}SO_{4} = NO + CuSO_{4} + Na_{2}SO_{4} + H_{2}O

**The equation is balanced using the following steps by the hit-and-trial method.**

**NaNO _{3} + Cu + H_{2}SO_{4} = NO + CuSO_{4} + Na_{2}SO_{4} + H_{2}O **

**First, we will compare the reactants’ atoms with the product’s atoms.**

Reactant Side | Product Side |
---|---|

1 Sodium unit from Sodium Nitrate | 1 Sodium unit from Sodium Nitrate |

1 Nitrogen unit from Sodium Nitrate | 1 Nitrogen unit from Nitrogen Oxide |

3 Oxygen units from Sodium Nitrate | 2 Oxygen units from Nitrogen Oxide and Water |

1 Copper unit | 1 Copper unit |

2 Hydrogen units from Sulphuric Acid | 2 Hydrogen units from Sulphuric Acid |

1 Sulphate unit from Sulphuric Acid | 2 Sulphate units from Sodium Sulphate and copper sulphate |

**Comparison of the atoms of the reactant side with the product side**

**Now, we will put the coefficients such that the atoms of reactants are equal to lớn the atoms of the products.****Thus, the balanced equation is:****2NaNO**_{3}+ 3Cu + 4H_{2}SO_{4}= 2NO + 3CuSO_{4}+ Na_{2}SO_{4}+ 4H_{2}O

## How to lớn Balance NaNO_{2} + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} = NaNO_{3} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + H_{2}O

**The equation is balanced using the following steps by the hit-and-trial method.**

**NaNO _{2} + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} = NaNO_{3} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + H_{2}O **

**Foremost, we will compare the atoms of reactants atoms with the atoms of the product.**

Reactant Side | Product Side |
---|---|

1 Sodium unit from Sodium Nitrite | 1 Sodium unit from Sodium Nitrate |

1 Nitrogen unit from Sodium Nitrite | 1 Nitrogen unit from Sodium Nitrate |

9 Oxygen units from Sodium Nitrite and Potassium dichromate | 4 Oxygen units from Sodium Nitrate and Water |

2 Potassium units from Potassium dichromate | 2 Potassium units from Potassium sulphate |

2 Chromium units from Potassium dichromate | 2 Chromium units from Chromium Sulphate |

2 Hydrogen units from Sulphuric Acid | 2 Hydrogen units from Water |

1 Sulphate unit from Sulphuric Acid | 4 Sulphate units from Chromium Sulphate and Potassium Sulphate |

**Comparison of the atoms of the reactant side with the product side**

**Now, we will mark the coefficients ví that the atoms of reactants are equivalent to lớn the atoms of the products.****Thus, the balanced equation is:****NaNO**_{2}+ 3K_{2}Cr_{2}O_{7}+ 4H_{2}SO_{4}= 3NaNO_{3}+ Cr_{2}(SO_{4})_{3}+ K_{2}SO_{4}+ 4H_{2}O

## How to lớn Balance NaNO_{3} + KI + H_{2}SO_{4} = I_{2} + NO + K_{2}SO_{4} + Na_{2}SO_{4} + H_{2}O

**The equation is balanced using the following steps by the hit-and-trial method.**

**NaNO _{3} + KI + H_{2}SO_{4} = I_{2} + NO + K_{2}SO_{4} + Na_{2}SO_{4} + H_{2}O **

**To balance the equation, we will equate the reactant atom with the product atom.**

Reactant Side | Product Side |
---|---|

1 Sodium unit from Sodium Nitrate | 2 Sodium units from Sodium Sulphate |

1 Nitrogen unit from Sodium Nitrate | 1 Nitrogen unit from Nitrogen Oxide |

3 Oxygen units from Sodium Nitrate | 2 Oxygen units from Nitrogen Oxide and Water |

1 Potassium unit from Potassium Iodide | 2 Potassium units from Potassium Sulphate |

1 Iodide unit from Potassium Iodide | 2 Iodine units from Iodide |

2 Hydrogen units from Sulphuric Acid | 2 Hydrogen units from Water |

1 Sulphate unit from Sulphuric Acid | 2 Sulphate units from Potassium Sulphate and Sodium Sulphate |

**Comparison of the atoms of the reactant side with the product side**

**Now, we will put the coefficients such that the atoms of the reactants equal the product atoms.****Thus, the balanced equation is:****2NaNO**_{3}+ 6KI + 4H_{2}SO_{4}= 3I_{2}+ 2NO + 3K_{2}SO_{4}+ Na_{2}SO_{4}+ 4H_{2}O

## How to lớn Balance NaNO_{3} + KMnO_{4} + H_{2}SO_{4} = NO_{2} + K_{2}SO_{4} + Na_{2}SO_{4} + MnSO_{4} + H_{2}O

**The equation is balanced using the following steps by the hit-and-trial method.**

**NaNO _{3} + KMnO_{4} + H_{2}SO_{4} = NO_{2} + K_{2}SO_{4} + Na_{2}SO_{4} + MnSO_{4} + H_{2}O **

**To balance the equation, we will equate the reactant atom with the product atom.**

Reactant Side | Product Side |
---|---|

1 Sodium unit from Sodium Nitrate | 2 Sodium units from Sodium Sulphate |

1 Nitrogen unit from Sodium Nitrate | 1 Nitrogen unit from Nitrogen dioxide |

7 Oxygen units from Sodium Nitrate and Potassium Permanganate | 3 Oxygen units from Nitrogen dioxide and Water |

1 Potassium unit Potassium Permanganate | 2 Potassium units Potassium Sulphate |

1 Manganese unit from Potassium Permanganate | 1 Manganese unit from Manganese Sulphate |

2 Hydrogen units from Sulphuric Acid | 2 Hydrogen units from Water |

1 Sulphate unit from Sulphuric Acid | 3 Sulphate units from Potassium Sulphate, Manganese Sulphate and Sodium Sulphate |

**Comparison of the atoms of the reactant side with the product side**

**Now, we will put the coefficients ví that the atoms of the reactants are equivalent to lớn the product atoms.****Thus, the balanced equation is:****10NaNO**_{3}+ 2KMnO_{4}+ 8H_{2}SO_{4}= 15NO_{2}+ K_{2}SO_{4}+ 5Na_{2}SO_{4}+ 2MnSO_{4}+ 8H_{2}O

#### Conclusion

Sulphuric acid reacts with Sodium Nitrate to lớn give a salt metathesis reaction yielding sodium sulphate and nitric acid. It is a reversible reaction; no oxidation or reduction occurs during the reaction, i.e., it is not a redox reaction.

Read more facts on H2SO4:

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H2SO4 + KClO3 H2SO4 + NaH H2SO4 + NaOCl H2SO4 + K2S H2SO4 + MnO2 H2SO4 + HCOOH H2SO4 + Mn2O7 H2SO4 + Mg H2SO4 + Na2CO3 H2SO4 + Sr(NO3)2 H2SO4 + MnS H2SO4 + NaHSO3 H2SO4 + CaCO3 H2SO4 + CH3COONa H2SO4 + Sn H2SO4 + Al2O3 H2SO4 + SO3 H2SO4 + H2O H2SO4 + Fe2S3 H2SO4 + NH4OH HCl + H2SO4 H2SO4 + FeCl2 H2SO4 + ZnCl2 H2SO4 + Al2(SO3)3 | H2SO4 + KOH H2SO4 + CH3CH2OH H2SO4 + Li2O H2SO4 + K2Cr2O7 H2SO4 + NaOH H2SO4+ Ag H2SO4 + Mn3O4 H2SO4 + NaH2PO4 H2SO4 + Sr H2SO4 + Zn H2SO4-HG2(NO3)2 H2SO4 + Pb(NO3)2 H2SO4 + Na H2SO4 + Ag2S H2SO4 + BaCO3 H2SO4 + PbCO3 H2SO4 + Sr(OH)2 H2SO4 +Mg3N2 H2SO4 + LiOH H2SO4 + Cl2 H2SO4 + AlCl3 H2SO4 + AlPO4 H2SO4 + Li2SO3 | H2SO4 + Fe(OH)3 H2SO4 + Al(OH)3 H2SO4 + NaI H2SO4 + K2CO3 H2SO4 + NaNO3 H2SO4 + CuO H2SO4 + Fe2O3 H2SO4 + AgNO3 H2SO4 + Al H2SO4 + K2SO4 H2SO4-HGO H2SO4 + MnCO3 H2SO4 + K2SO3 H2SO4 + PbCl2 H2SO4 + P4O10 H2SO4 + NaHCO3 H2SO4 + O3 H2SO4 + Ca(OH)2 H2SO4 + Be(OH)2 H2SO4 + BaSO3 H2SO4 + Fe3O4 H2SO4 + Zn(OH)2 H2SO4 + KI | H2SO4 + KMnO4 H2SO4 + CH3NH2 H2SO4 + CH3COOH H2SO4 + Pb H2SO4 + CH3OH H2SO4 + Fe2(CO3)3 H2SO4 + Li2CO3 H2SO4 + MgO H2SO4 + Na2O H2SO4 + F2 H2SO4 + Zn(NO3)2 H2SO4 + Ca H2SO4 + K2O H2SO4 + Mg(OH)2 H2SO4+NaF H2SO4 + Sb2S3 H2SO4 + NH4NO3 H2SO4 + AlBr3 H2SO4 + CsOH H2SO4 + KBr H2SO4 + Na2HPO4 H2SO4 + Na2S2O3 H2SO4 + BeO | H2SO4 + Fe H2SO4 + HCOONa H2SO4 + Cu H2SO4 + PbS H2SO4 + P2O5 H2SO4 + CuCO3 H2SO4 + Li H2SO4 + K2CrO4 H2SO4 + NaCl H2SO4 + Ag2O H2SO4 +Mg2Si H2SO4 + Mn(OH)2 H2SO4+ NACLO2 H2SO4 + K H2SO4 + CaCl2 H2SO4 + Li2S H2SO4 + H2O2 H2SO4 + CuS H2SO4 + Li3PO4 H2SO4 + Be H2SO4 + Na2S H2SO4 + Fe3O4 H2SO4 + As2S3 |

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