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Chapter 21 Section 1

Chapter 21 Section 1. neutralization reaction - when an acid and a base react in an aqueous solution to produce a salt and water *all neutralization reactions are double-replacement ex- RS + TU  TS + RU *must be balanced acid + base  salt + water

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Chapter 21 Section 1

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  1. Chapter 21 Section 1 neutralization reaction- when an acid and a base react in an aqueous solution to produce a salt and water *all neutralization reactions are double-replacement ex- RS + TU  TS + RU *must be balanced acid + base  salt + water salt- compound consisting of an anion from an acid and a cation from a base

  2. -The concentration of acid (or base) in a solution can be determined by a neutralization reaction -use acid-base indicators ex- phenolphthalein turns from clear to pink as pH changes from acidic to basic pg 616 figure 21.3 pg 617 figure 21.4

  3. Steps of a Neutralization Reaction -Read and record from page 617 (#1, 2 and 3) standard solution- solution of known concentration *added using a buret titration- process of adding a known amount of solution of known concentration to determine the concentration of another solution end point- point at which the indicator changes color *solution is neutralized

  4. equivalent (equiv)- the amount of acid (or base) that will give one mole of H+ or OH- ions ex- 1mol of HCℓ = 1 equiv of HCℓ 1 mol of H2SO4 = 2 equiv of H2SO4 1 mol of NaOH = 1 equiv of NaOH 1 mol of Ca(OH)2 = 2 equiv of Ca(OH)2 *1 mol of HCℓ will neutralize 1 mol of NaOH *1 mol of HCℓ will not neutralize 1 mol of Ca(OH)2 b/c you need 2 mol

  5. -in a neutralization reaction, the equiv of acid must equal the equiv of base gram equivalent mass- the mass of 1 equiv of a substance

  6. Normality N= equiv/L equiv= (N)(L) *Normality and molarity are equal for acids and bases that give 1 equiv/mol of H+ and OH- ex- 1M HCℓ = 1N HCℓ 1 M NaOH = 1N NaOH *but 1M Ca(OH)2 = 2N Ca(OH)2

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