The Reaction between Potassium Hydroxide and Sulfuric acid

 Answer the question in Figure 1. 

Figure 1: The Reaction between potassium hydroxide and sulfuric acid

First, students must write a balanced chemical equation between sulfuric and potassium hydroxide to answer this question. Potassium hydroxide is a base, and when a base reacts with an acid, salt, and water is produced. Therefore, the general equation of the reaction between sulfuric acid and the base becomes: 

Base + Acid Salt + water                                                  Equation 1 

Figure 2: A conical flask for titrations

Now, by inserting potassium hydroxide and sulfuric acid in Equation 1, the equation leads to Equation 2.  

KOH (aq) + H2SO4 (aq) K2SO4 (aq) + H2O (l)                 Equation 2 (Unbalanced equation) 

On the left side of Equation 2 are one mole of potassium, three moles of hydrogen, one mole of sulfur, and five moles of oxygen in reactants. On the right hand of the same equation are two moles of potassium, one mole of sulfur, five moles of oxygen, and two moles of hydrogen on the product side. For this reason, Equation 2 is not balanced. The number of potassium and hydrogen moles is not the same on both sides. Therefore, they must be balanced by introducing appropriate numbers, which results in Equation 3. 

2KOH (aq) + H2SO4 (aq) K2SO4 (aq) + 2H2O (l)             Equation 3 

Introducing stoichiometric coefficients two before potassium hydroxide on the left-hand side and two before water balances the equation. Based on the equation, 2:1 is the mole ratio between the potassium hydroxide and sulfuric acid, which is now applied to the volumetric analysis. 

The variables (n, c, V) are used in volumetric analysis. n denotes the number of moles, c is the concentration(mol/L or mol/dm3), and V is the volume(dm3 or mL).Now, by using the relationship concentration(mol/L) and volume(Ml) in the dilution formula(M1V1=M2V2), the required volume can be determined. 

In the dilution calculation formula, M1 (in molar) and V1 (mL) are the concentration of potassium hydroxide. On the other hand, M2 (in molar) and V2 (mL) are the concentration and volume of sulfuric acid. Furthermore, 2 moles of potassium hydroxide require one mole of sulfuric acid to form the product. Hence, to obtain 20cm3 of 0.5M of sulfuric, we must multiply the sulfuric acid by two (the stoichiometric coefficient). Potassium hydroxide is an excess reactant, while sulfuric is a limiting reactant. 

(0.8M*V1) =2(0.5M*20cm3)

V1= (2(0.5M*20cm3))/0.8M

V1= 25cm3 

 

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