# Neutralization Chapter 21.

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Neutralization Chapter 21

Neutralization Reactions
A reaction between a strong acid and a strong base in an aqueous solution that produces a salt and water- a salt is an ionic compound These are double replacement reactions Used to produce pure salts HCl KOH KCl + H2O The water is evaporated leaving behind the salt KCl

Titrations- used to calculate the unknown concentration of an acid or base when you know the concentration of the other. Standard solution- solution of known concentration End Point- the point in the titration where the indicator changes color

From a balanced equation, you can determine how many moles of a substance is needed to react with another substance to be neutralized: EXAMPLE: How many moles of sulfuric acid are needed to neutralize 0.5 mol of sodium hydroxide? H2SO NaOH Na2SO H2O Using stoichiometry: 0.5 mol NaOH x 1 mol H2SO4 2 mol NaOH = 0.25 mol H2SO4 Do 1 & 2 page 616

Stoichiometry can be used to determine the strength of an unknown acid or base:
Titrations: Steps: A measured volume of unknown concentration acid or base is added to a flask Several drops of the indicator is added to the solution Measured volume of a base or acid of known concentration are mixed into the flask until the indicator just barely changes color

EXAMPLE: A 25 mL solution of H2SO4 is completely neutralized by 18 mL of 1.0 M NaOH using phenolphthalein as an indicator. What is the concentration of the H2SO4solution? H2SO4 + 2NaOH Na2SO4 + 2H2O L NaOH x 1.0 mol NaOH x 1 mol H2SO4 1 L NaOH 2 mol NaOH = mol H2SO L = 0.36 M

Equivalents- the amount of acid or base that will give 1 mole of hydrogen or hydroxide Examples: HNO3 H NO3- 1 mole of HNO3 gives 1 mole H+ 1 mole HNO3 is 1 equivalent of HNO3 H2SO4 2H SO42- 1 mole of H2SO4 is 2 equivalents of H2SO4

Gram Equivalent Mass- the mass of one equivalent of a substance Molar Mass = gram equiv. mass # equiv/mol Example: HCl = 36.5/1 equiv/mol = 36.5 g/equiv H2SO4 = 98 g/mol/2 equiv/mol = 49 g/equiv

Normality- concentration of a solution expressed as the number of equivalents of solute in 1 liter of solution N = equiv / L Equivalence point- the point of neutralization in a titration The number of equivalents of acid and base are equal at this point so: NA x VA = NB x VB