PbCl_{2} is a white solid inorganic compound under ambient conditions, and H_{2}SO_{4} is a strong acid. Let us see the reaction of H_{2}SO_{4} and PbCl_{2}.

**PbCl _{2} naturally occurs in the ore form, Cotunnite, and Sulfuric acid, a colorless and clear solution. PbCl_{2} is poorly soluble in water. H_{2}SO_{4} known in antiquity as oil of vitriol, is a mineral acid**.

**Pure sulfuric acid does not exist naturally on Earth due to its strong affinity to water vapor.**

This article will discuss some facts about the reaction, like the type of reaction and net ionic equation.

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

_{2}SO

_{4}and PbCl

_{2}

**Lead sulfate and hydrogen chloride are product of the H_{2}SO_{4} +**

**PbCl**_{2}**reaction**.

**H _{2}SO_{4} + PbCl_{2}** =

**Pb(SO)**

_{4}+ 2 HCl**What type of reaction is H**_{2}SO_{4} + PbCl_{2}

_{2}SO

_{4}+ PbCl

_{2}

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

**is a precipitate exchange reaction**.

**How to balance H**_{2}SO_{4} + PbCl_{2}

_{2}SO

_{4}+ PbCl

_{2}

**The reaction** **H _{2}SO_{4} + PbCl_{2}**

**is**

**balanced**

**using**

**the**

**following**

**steps:**

**H _{2}SO_{4} + PbCl_{2}** =

**Pb(SO)**

_{4}+ 2 HCl**The first thing we check is the number of atoms on the reactant and product sides.****After that, if the number of atoms is not the same, then we have done multiplication of the smallest figure according to the formula**.**In the above equation, the H atom and Cl atoms number are not the same****In this next step, we should multiply HCl by 2.****Thus the balance equation is****H**=_{2}SO_{4}+ PbCl_{2}**Pb(SO)**_{4}+ 2 HCl

**H**_{2}SO_{4} + PbCl_{2} titration

_{2}SO

_{4}+ PbCl

_{2}titration

**Titration is not possible in the reaction H _{2}SO_{4 }+ PbCl_{2} because H_{2}SO_{4} is an acid, and when it reacts with PbCl_{2}, it forms a precipitate of Pb(SO_{4}).**

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

_{2}SO

_{4}+ PbCl

_{2}net ionic equation

**The net ionic equation for the reaction H_{2}SO_{4 }+ PbCl_{2} is**

**PbCl _{2}(s) + SO_{4}^{2-}(aq) = PbSO_{4}(s) + 2Cl^{–}(aq)**

**To derive the net ionic equation, the following steps are followed**

**First, the soluble ionic compounds can dissociate into ions**.**PbCl**_{2}() + 2H*s*^{+}() + SO*aq*_{4}^{2-}() = PbSO*aq*_{4}() + 2H*s*^{+}() + 2Cl*aq*^{–}()*aq***Spector ions 2H**^{+}**is cancelled out.****Thus the net ionic equation****is****PbCl**_{2}() + SO*s*_{4}^{2-}() = PbSO*aq*_{4}() + 2Cl*s*^{–}()*aq*

**H**_{2}SO_{4} + PbCl_{2} conjugate pairs

_{2}SO

_{4}+ PbCl

_{2}conjugate pairs

**The Conjugate pairs of** **H _{2}SO_{4 }+ PbCl_{2}**

**in below**

**:**

**H**_{2}SO_{4 }+ PbCl_{2}**does not form conjugate pairs because PbSO**_{4}is in precipitate form.**conjugated****base of H**_{2}SO_{4}is HSO_{4}^{–}.

**H**_{2}SO_{4} + PbCl_{2} intermolecular forces

_{2}SO

_{4}+ PbCl

_{2}intermolecular forces

**Dipole-dipole interactions and strong hydrogen bonding are present as intermolecular forces****in****H**._{2}SO_{4}**weak London dispersion forces is present in****PbCl**._{2}

**H**_{2}SO_{4} + PbCl_{2} reaction enthalpy

_{2}SO

_{4}+ PbCl

_{2}reaction enthalpy

**The reaction enthalpy of H_{2}SO_{4 }+ PbCl_{2} reaction is**

**14.21 KJ**.

**Is H**_{2}SO_{4} + PbCl_{2} a buffer solution?

_{2}SO

_{4}+ PbCl

_{2}a buffer solution?

**H _{2}SO_{4 }+ PbCl_{2}** reaction can not form a buffer solution because

**H**is a strong acid.

_{2}SO_{4}**Is H**_{2}SO_{4} + PbCl_{2} a complete reaction?

_{2}SO

_{4}+ PbCl

_{2}a complete reaction?

**H _{2}SO_{4 }+ PbCl_{2} **is a complete reaction because

**lead sulfate and**hydrogen chloride are formed in reaction.

**Is H**_{2}SO_{4} + PbCl_{2} an exothermic or endothermic reaction?

_{2}SO

_{4}+ PbCl

_{2}an exothermic or endothermic reaction?

**H _{2}SO_{4 }+ PbCl_{2}** is an

**endothermic reaction because ΣΔH**.

^{°}_{f}(products) is greater than ΣΔH^{°}_{f}(reactants)**Is ****H**_{2}SO_{4} + PbCl_{2} a redox reaction?

**H**a redox reaction?

_{2}SO_{4}+ PbCl_{2}**H _{2}SO_{4} + PbCL_{2} is not a redox reaction.**

**Is ****H**_{2}SO_{4} + PbCl_{2} a precipitation reaction?

**H**a precipitation reaction?

_{2}SO_{4}+ PbCl_{2}**H _{2}SO_{4 }+ PbCl_{2}** is a precipitation reaction as PbSO

_{4}produces precipitate.

**Is ****H**_{2}SO_{4} + PbCl_{2} reversible or irreversible reaction?

**reversible or irreversible reaction?**

**H**_{2}SO_{4}+ PbCl_{2}**H _{2}SO_{4 }+ Pb**Cl

**is irreversible reaction because the product of the reaction is a residue, so it is an irreversible reaction.**

_{2}**Is ****H**_{2}SO_{4} + PbCl_{2} displacement reaction?

**displacement reaction?**

**H**_{2}SO_{4}+ PbCl_{2}** H_{2}SO_{4 }+ PbCl_{2} is a double displacement reaction because H replaces Pb from Pb(NO_{3})_{2} and formed HNO_{3} , and Pb replaces H from H_{2}SO_{4} and formed PbSO_{4}** .

**Conclusion**

From this article, we understand that this is a double displacement, irreversible, and endothermic reaction, which means it needs some energy to do the reaction. We can use this reaction to produce HCl.