CsOH is an alkali hydroxide with pKa of 15.76 that shows its strong basic nature. H_{2}SO_{4} is oily liquid soluble in H_{2}O with the release of heat. Let us see how it reacts with H_{2}SO_{4}.

**CsOH is water soluble and very reactive as it donates OH ^{–}easily. It acts as a base (pH=10.86) and reacts with acid at a high reaction rate. CsOH is used in fuel cells as an electrolyte. H_{2}SO_{4}, an oxidizing agent, is known as the oil of vitriol**.

**It is a strong acid due to its ease of H**

^{+}emit.We will discuss here some important facts about the H_{2}SO4 + CsOH reaction, like the products of the reaction, the enthalpy, intermolecular forces between their compounds, and the mechanism behind the reaction.

## What is the product of H_{2}SO_{4 }+ CsOH?

**Cesium sulfate (Cs_{2}SO**

_{4}

**) and water (H**

_{2}

**O)**are formed in the reaction CsOH + H

_{2}SO

_{4}in which CsSO

_{4 }is the major product.

**2CsOH + H _{2}SO_{4} → Cs_{2}SO_{4} + 2H_{2}O**

## What type of reaction is H_{2}SO_{4 }+ CsOH

**H _{2}SO_{4 }+ CsOH is precipitation, Acid-base, Double displacement (Metathesis), and Endothermic Reaction.**

## How to balance H_{2}SO_{4 }+ CsOH

**The unbalanced molecular equation for H _{2}SO_{4 }+ CsOH is. **

**CsOH (aq) + H _{2}SO_{4} (aq) = Cs_{2}SO_{4} (aq) + H_{2}O (l)**

**To balance this equation, we should follow the steps given below: **

**Here, the number of Cs and H atoms is not the same on both sides of the reaction. So we will multiply these atoms with some coefficients so that they become equal.****The total number of Cs and H atoms,****on the reactant side**is 1 and 3 respectively while on the product side is 2 and 2**respectively**.**So, we multiply the CsOH with a coefficient of 2 on the reactant side and H**_{2}O with a coefficient of 2 also on the product side. So the number of Cs and H atoms becomes 2 and 4 on both sides of the reaction, respectively.**Finally, the balanced equation****is****2CsOH (aq) + H**_{2}SO_{4}(aq) = Cs_{2}SO_{4}(aq) + 2H_{2}O (l)

## H_{2}SO_{4 }+ CsOH titration

**Quantitative estimation of H_{2}SO_{4} is done by performing the titration of the H_{2}SO_{4} solution against a basic solution of CsOH. For titration, the**

**following procedure is carried out**.

**Apparatus used**

**A burette, Pipette, measuring flask, glass funnel, clamp stand, measuring cylinder, volumetric flask, and beakers are required for this titration.**

**Indicator**

**Phenolphthalein is used as an indicator for this titration because this reaction is an acid-base reaction. **

**Procedure**

**A standard amount of CsOH solution is filled into the burette and ensures that the interior portion of the burette is coated with a thin layer of CsOH solution.****The dilute solution of H**_{2}SO_{4 }is taken into a conical flask with the help of a pipette and a few drops of phenolphthalein are added to this acidic solution.**CsOH is released****very carefully****in a dropwise manner into the conical flask and keeps constant shaking of the flask gently as CsOH starts reacting with the acid solution.****Add CsOH into the H**_{2}SO_{4}solution until the pink color comes. The procedure is repeated at least three times until a constant endpoint comes.**After the successful titration, the strength of sulfuric acid and the quantity of sulfate ions will be measured by the formula V**_{1}N_{1 }= V_{2}N_{2}.

## H_{2}SO_{4 }+ CsOH net ionic equation

**The net chemical ionic equation of H_{2}SO_{4 }+ CsOH is**:

**2H ^{+}(aq) + 2OH^{–} (aq) = H_{2}O (l) **

**To get this ionic equation we should follow the steps given below**.

**Write the solubility equation for****CsOH + H**_{2}SO_{4}**by****labeling the state or phase (s, l, g or aq) of each substance in the balanced equation of****H**_{2}SO_{4 }+ CsOH**2CsOH (aq) + H**_{2}SO_{4}(aq) = Cs_{2}SO_{4}(aq) + 2H_{2}O (l)**Break all aquatic soluble ionic substances into their corresponding ions to get the****complete ionic equation****2Cs**^{+}(aq) +2OH^{–}(aq) + 2H^{+ }(aq) + SO_{4}^{2-}(aq) = H_{2}O (l) + 2Cs^{+ }(aq) +**SO**(aq)_{4}^{2-}**Then remove spectator ions (****SO**and Cs_{4}^{2-}^{+}) from L.H.S and R.H.S of the complete ionic equation.**Finally, the net ionic equation is:****2H**^{+}(aq) +**2OH**(aq) =^{–}**H**_{2}O (l)

## H_{2}SO_{4 }+ CsOH conjugate pairs

**The conjugate pairs of reactant and product in the H_{2}SO_{4 }+ CsOH**

**reaction is**

**as follows:**

**The conjugate base of H**_{2}SO_{4}is HSO_{4}^{–}**The conjugate base of H**_{2}O is OH^{–}**Cs**_{2}SO_{4}and CsOH do not have any conjugate pairs (acid and base) because hydrogen is not present in both these elements so it can release in the proton form.

## H_{2}SO_{4 }+ CsOH intermolecular forces

**The reactants and products in H _{2}SO_{4 }+ CsOH reaction are held with each other by following intermolecular forces**.

**Van der Waals dispersion forces, dipole-dipole interactions, and hydrogen bonding****are exhibited in the H**_{2}SO_{4}**molecules****as****H**molecules are highly polar with a 2.725 D dipole moment._{2}SO_{4}**Electrostatic force or Coulomb force is present in the Cs**_{2}SO_{4}molecules due to their ionic nature.**Hydrogen bonding exists in the H**_{2}O molecules.

## H_{2}SO_{4 }+ CsOH reaction enthalpy

**The total enthalpy for the reaction H _{2}SO_{4 }+ CsOH is 368.66 kJ/mol. The enthalpy of the formation is calculated using the values given below**.

Compound | Standard Formation Enthalpy (Δ_{f}H°(Kj/mol) ) |
---|---|

CsOH | -416 |

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

Cs_{2}SO_{4} | -1443 |

H_{2}O | -285.83 |

**Standard Formation Enthalpy of Reactants and products**

**ΔH°**_{f}= ΣΔH°_{f}(products) – ΣΔH°_{f}(reactants)**(kJ/mol)****ΔH**_{f}= [2*(-416) – 814) – (-1443 – 2*(285.83))**kJ/mol****ΔH**_{f}= 368.66 kJ/mol

## H_{2}SO_{4 }+ CsOH a buffer solution

**The reaction H _{2}SO_{4 }+ CsOH provides a buffer solution of Cs_{2}SO_{4} and water which resist the change in its pH value if we add acid or base to it. **

## Is H_{2}SO_{4 }+ CsOH a complete reaction

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

**is a complete reaction because in this reaction**

**completely neutralizes****H**_{2}SO_{4}**CsOH****molecules in its corresponding salt Cs**

_{2}SO_{4}.## Is H_{2}SO_{4 }+ CsOH an exothermic or endothermic reaction

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

**is an endothermic reaction as the total enthalpy change for the reaction is 368.66 kJ/mol which is positive, w**

**here the positive sign explains the following facts about the reaction**.

**The reactants H**_{2}SO_{4}and CsOH absorb the heat from the surrounding atmosphere which lowers the temperature of the surroundings**and ultimately the reaction system gets cool.****Products ( CsSO**_{4}and H_{2}O ) exhibit higher energy than reactants and become unstable during the reaction.

## Is H_{2}SO_{4 }+ CsOH a redox reaction

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

**is not a redox reaction because the oxidation states of each element remain the same even after the completion of the reaction.**

**Cs ^{+1 }O ^{-2} H ^{+1} + H_{2}^{+1} S ^{+6} O_{4 }^{-2} → Cs_{2 }^{+1} S ^{+6} O_{4 }^{-2} + 2 H_{2}^{+1} O ^{-2}**

## Is H_{2}SO_{4 }+ CsOH a precipitation reaction

**H _{2}SO_{4 }+ CsOH is not a precipitation reaction because all the products (H_{2}O and Cs_{2}SO_{4}) formed are in liquid and aqueous states. **

## Is H_{2}SO_{4 }+ CsOH reversible or irreversible reaction

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

**is an irreversible reaction b**

**ecause, at the same reaction conditions, it is not possible to get reactants back from the products.**

## Is H_{2}SO_{4 }+ CsOH displacement reaction

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

**is a double displacement reaction because**

**Cs from Cs(OH) and H from**

**H**displace each other’s positions to form different products-_{2}SO_{4}**Cs**and H_{2}SO_{4}_{2}O.#### Conclusion

H_{2}SO_{4 }+ CsOH is an acid-base reaction in which H_{2}SO_{4} acts as a strong acid and CsOH as a base. In this reaction, 368.66 kJ/mol energy is consumed by the reactants to form cesium hydroxide. During this reaction, the transfer of electrons does not take place, hence this reaction does not follow to redox process.

Hi, I am Kavita Singhal, Ph.D. in Chemical Sciences. My subject area of interest is Physical and Inorganic Chemistry with special emphasis on Electrochemistry, Polymer Chemistry, Nano Chemistry, Corrosion Study, Cyclic voltammetry, Supercapacitance, and Organometallic Chemistry.

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