In a chemical reaction chemical symbols are used to denote reactant and product. Let us discuss about the reaction of H_{2}SO_{4 } and CH_{3}COONa.

**Sulphuric acid or H _{2}SO_{4} is a strong acid and sodium acetate, CH_{3}COONa is a sodium salt if acetic acid. Both dissociate in water to give respective ions and are soluble in water. Sodium acetate is a colorless substance which is deliquescent in nature. It’s crystal structure is monoclinic.**

Sodium acetate is a buffering agent with molar mass 82.034 g/mol. Let us discuss about H_{2}SO_{4}+ CH_{3}COONa with all details.

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

_{2}SO

_{4}and CH

_{3}COONa

**H _{2}SO_{4} reacts with CH_{3}COONa to give acetic acid and sodium hydrogen sulphate.**

**H _{2}SO_{4} + CH_{3}COONa ——–> CH_{3}COOH + NaHSO_{4}**

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

_{2}SO

_{4}+ CH

_{3}COONa

**H _{2}SO_{4} + CH_{3}COONa is a displacement reaction. In this reaction sodium is displaced from sodium acetate by hydrogen of H_{2}SO_{4} to yield the products.**

**How to balance H**_{2}SO_{4} + CH_{3}COONa

_{2}SO

_{4}+ CH

_{3}COONa

**Balancing a chemical reaction is needed and it gives information about the number of molecules reacted. Following is the reaction balancing steps.**

**Assigning certain coefficients to both reactants and products**

**a****H**+ b_{2}SO_{4}**CH**—–> c_{3}COONa**CH**+ d NaHSO_{3}COOH_{4}

**An equation is made with the above equation**

**H= 2a= 3b =4c = d, S =a =d, O =4a= 2b = 2c= 4d, C =2b =2c, Na = b= d**

**2a +3b= 4c +d, a=d, 4a+ 2b=2c+ 4d, 2b=2c, b =d**

**Solve the above equation using gauss elimination method**

**a =1, b= 1, c= 1, d= 1**

**The balanced equation of H**_{2}SO_{4}+ CH_{3}COONa is

**H**_{2}SO_{4}+ CH_{3}COONa —–> CH_{3}COOH + NaHSO_{4 }

**H**_{2}SO_{4} + CH_{3}COONa titration

_{2}SO

_{4}+ CH

_{3}COONa titration

**H _{2}SO_{4} + CH_{3}COONa titration can be performed. It is because one is strong acid and other one is a weak base. So they can be titrated against to get the endpoint sharply.**

**Apparatus**

**Conical flask, pipette, burette, H _{2}SO_{4} and CH_{3}COONa solution.**

**Indicator**

**Methyl orange is used because we are titrating strong acid and weak base**

**Procedure**

**20 ml of sodium acetate solution is pipetted into a conical flask and burette is filled with H**_{2}SO_{4}.

**Methyl orange indicator is added into the solution in conical flask until it turns into a yellow color.**

**Titrate the solution against sulphuric acid in burette until the color changes from yellow to red color.**

**Note down the burette reading and do the respective calculations.**

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

_{2}SO

_{4}+ CH

_{3}COONa net ionic equation

**The net ionic equation of H _{2}SO_{4} and CH_{3}COONa is,**

**2H ^{+} + SO^{2-}_{4 }+ CH_{3}COO^{– } ——> 2CH_{3}COOH + HSO^{–}_{4 }.**

**H**_{2}SO_{4} and CH_{3}COONa conjugate Pairs

_{2}SO

_{4}and CH

_{3}COONa conjugate Pairs

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

**The conjugate pairs of H**_{2}SO_{4}and CH_{3}COONa is HSO^{–}_{4 }and CH_{3}COOH respectively.**Both are acids so one of the proton is eliminated to get its conjugate base.**

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

_{2}SO

_{4}and CH

_{3}COONa intermolecular forces

**The intermolecular forces exist in between H _{2}SO_{4} + CH_{3}COONa is **

**In H**_{2}SO_{4}there is ionic force of attraction can be seen in between H^{+}ion and SO_{4}^{2-}.**In CH**_{3}COONa ionic force between sodium and acetate, dipole dipole interaction between carbon and oxygen and pure covalent bonds between carbon and hydrogen.

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

_{2}SO

_{4}+ CH

_{3}COONa reaction enthalpy

**The reaction enthalpy of H _{2}SO_{4} + CH3COONa is positive. The enthalpy of formation of H_{2}SO_{4}, CH_{3}COOH, and CH_{3}COONa is -814, -491, and -24.6 kJ/mol respectively. **

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

_{2}SO

_{4}+ CH

_{3}COONa a buffer solution

**H _{2}SO_{4} + CH_{3}COONa is not a buffer solution. A buffer solution is made by mixing a weak acid or weak base with its conjugate base or acid respectively. Here H_{2}SO_{4} a strong acid. But CH_{3}COONa is a weak acid it’s buffer can be prepared by mixing with its conjugate acid like CH_{3}COOH to get its buffer solution.**

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

_{2}SO

_{4}+ CH

_{3}COONa a complete reaction

**H _{2}SO_{4} + CH_{3}COONa is a complete reaction. The reaction takes place at room temperature to yield acetic acid and sodium hydrogen sulphate. Here sulphuric acid displaces weaker ions from its salts.**

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

_{2}SO

_{4}+ CH

_{3}COONa an exothermic or endothermic reaction

**H _{2}SO_{4} + CH_{3}COONa is an endothermic reaction. The reaction doesn’t liberate any amount of heat. So it is an endothermic reaction.**

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

_{2}SO

_{4}+ CH

_{3}COONa a redox reaction

**H _{2}SO_{4} + CH_{3}COONa is not a redox reaction as the oxidation state of all the atoms are same before and after the reaction. **

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

_{2}SO

_{4}+ CH

_{3}COONa a precipitation reaction

**H _{2}SO_{4} + CH_{3}COONa is reaction doesn’t form any precipitate. Acetic acid and sodium hydrogen sulphate is the product formed here and both are liquid like nature.**

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

_{2}SO

_{4}+ CH

_{3}COONa reversible or irreversible reaction

**H _{2}SO_{4} + CH_{3}COONa is an irreversible reaction as they cant be reversed to yield it’s reactants**.

**Is H**_{2}SO_{4} + CH_{3}COONa a displacement reaction

_{2}SO

_{4}+ CH

_{3}COONa a displacement reaction

**H _{2}SO_{4} + CH_{3}COONa is a Displacement reaction because the weaker ion sodium from sodium acetate is replaced by hydrogen to form acetic acid.**

**Conclusion**

In this article H_{2}SO_{4} and CH_{3}COONa reacts to sodium hydrogen sulphate and acetic acid. This reaction is found to be a Displacement reaction which is not redox and irreversible while it is endothermic. The reaction does not form any precipitate.