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Solubility Rules and Precipitation Reactions in Chemistry

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Learn about solubility rules, metathesis reactions, and oxidation states in chemistry. Understand how to predict precipitate formation and identify spectator ions.

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Solubility Rules and Precipitation Reactions in Chemistry

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  1. Solubility Rules #2 Rule #1 Rule [Exceptions]

  2. Metathesis (Double Displacement) AX + BYAY + BX Acid-Base neutralization A=H+ X=anion of the acid BY=Base HCl + NaOHH2O + NaCl note: In the case of ammonia: AX + YAY + X HCl+NH3NH4+ +Cl-

  3. Precipitation Reactions (metathesis) AX + BYAY(s) + BX AX=Soluble Salt BY=Soluble Salt AY=Insoluble Solid (precipitate,ppt) BX=Soluble Salt AgNO3 + NaClAgCl(s) + NaNO3 Formula unit Ag++NO3- + Na+ + Cl-AgCl(s) + Na+ + NO3- Total Ionic Ag++ Cl-AgCl(s) Net Ionic Spectator Ions = nitrate ions and sodium ions

  4. Fe(NO3)3 + 3NaOH Fe(OH)3(s) + 3NaNO3 Fe3+ + 3NO3- + 3Na+ + 3OH- Fe(OH)3(s) + 3NO3- + 3Na+ Fe3+ + 3OH- Fe(OH)3(s)

  5. Will a precipitate form if we add solid Calcium Nitrate and solid Potassium Carbonate to water? Ca(NO3)2 Ca2+ + 2NO3- K2CO3 2K+ + CO32- Ca2+ NO3- K+ CO32- Ca(NO3)2 + K2CO3 CaCO3(s) + 2KNO3 Formula Unit Ca2+ + 2NO3- + 2K+ + CO32- CaCO3(s) + 2NO3- + 2K+ Total Ionic Ca2+ + CO32- CaCO3(s) Net Ionic Spectator ions: nitrate ions and potassium ions

  6. Oxidation State (Oxidation Number) 2Na(s) 2Na+(aq) + 2e- oxidation 1/2 reaction 2H2O + 2e- 2OH-(aq) + H2(g) reduction 1/2 reaction 2Na(s)+2H2O2NaOH(aq)+H2(g) redox reaction oxidation-loss of electrons reduction-gain of electrons reducing agent-compound that is oxidized (Na(s)) oxidizing agent-compound that is reduced (H2O) We need an accounting system-oxidation state

  7. Oxidation State Rules • The oxidation state of any atom in a free uncombined element is zero (Cl2, H2, O2, P4, Na(s) ... ) • The oxidation state of any monatomic ion is equal to the charge (Na+ ox. state = +1 ; Cl- ox. state = -1 ; Fe2+ ox. state = +2) • The sum of oxidation numbers of all atoms in a compound is zero • The sum of oxidation numbers of all atoms in an ion is equal to the charge of the ion (SO42- sum of ox. numbers = -2)

  8. 1 2 3 4 5 6 7 8

  9. -2 +1 0 -2 2Na(s) + 2H2O2NaOH(aq) + H2(g) +1 +1 0 H2O 2*(+1) + X = 0 X= -2 for O in H2O NaOH (+1) + (+1) + X = 0 X=-2 for O in NaOH OH-- (+1) + X = -1 X=-2 for O in OH- Oxidation-increase in the oxidation state (number) Na(s) 0 +1 Reduction-decrease in the oxidation state (number) H +1 0

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