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Module 1: Acid, Bases and Salts

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Base Neutralization Reaction

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NeutralisationIn this media, you will learn, about the neutralization reaction.Take a conical flask, with some dilute hydrochloric acid in it.Add, few drops of phenolphthalein indicator, to the acid solution.You will observe, that the solution is still colorless, due to the acidic solution.When we add, some sodium hydroxide solution to the acidic solution, the solution becomes basic in nature.Now, you will observe pink colored solution, because phenolphthalein gives pink color, in the basic solution.Add, some hydrochloric acid solution to it again.You will find, that the pink color starts disappearing, by itself due to the acidic nature of the solution.Thus, the solution becomes colorless, due to the acidic nature.Therefore, we find that, when an acid is mixed with a base, both the solutions neutralizeeach otherseffect.Hence the solution is neither acidic, nor basic in nature.You will also observe, rise in temperature of the solution, as it is actually due to the evolution of heat, from the neutralization reaction.Hence, neutralization reaction involves mixing of acids, and, bases in equal proportion to give, salt and water, with the evolution of heat. Thus, you have learned about, neutralization.
Acid-Base Neutralization Reaction CalculationThis video describes the calculations involved in acid-base neutralization reaction.Consider an example problem where 30 mL of 1 M HCl is mixed with 70 mL of 0.5 M NH3 with a base dissociation constant of 2 X 10-9. What is the pH of the solution?This reaction is identical to the reaction between strong acid (HCl) with a weak base (NH3).When hydrochloric acid mixes with a weak base like ammonia, we get ammonium ion and chloride ion.In this problem, the concentration of HCl = 1 Molar = 1 mol/L.The concentration of ammonia NH3 = 0.5 M = 0.5 mol/L.The volume of HCl used is 30 mL.The volume of ammonia NH3 used is 70 mL.The first step is to convert the concentration into number of moles of reactant.We get, the no. of moles of HCl (H+) used = 1 mol/L X 30 mL X (1 L)/(1000 mL) = 0.03 mol.The no. of moles of NH3 used = 0.5 mol/L X 70 mL X (1 L)/(1000 mL) = 0.035 mol.Using this information, we set up the ICE table as shown below.The reaction proceeds until all the H+ are used up and the products, a conjugate acid (NH4+) and remaining NH3 is at equilibrium. When acid and base is mixed, the sample volume = volume of acid + base used. So the new sample volume = 30 mL + 70 mL = 100 mL.The concentration of the products (the conjugate acid and NH3) after the reaction reaches equilibrium is now shown.Since we know the base dissociation constant Kb, we will use the Henderson Hasselbalch Equation for the pOH calculation and use this value to back calculate the pH of the solution.pOH of a solution=pK_b+log⁡〖 [Conjugate Acid,HB^+ ]/[Base,B] 〗In this equation pKb = negative logarithm of Kb = -log[Kb].Using the concentrations of conjugate acid and base after the reaction has reached equilibrium, we get pOH = -log(2 × 10^(-9) )+log ([0.3])/([0.05]).Solving the equation we get, pOH = 9.59.Using the equilibrium constant of water, we have pH + pOH = 14.Solving the equation, we get pH of the solution is 4.41.