HA OH K cb K w K a 1 x 10 14 1 x 10 5 1 x 10 9 Conjugate base H 3 O eg sodium

Ha oh k cb k w k a 1 x 10 14 1 x 10 5 1 x 10 9

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HA + OH-Kcb= Kw/Ka= (1 x 10-14)/(1 x 10-5) = 1 x 10-9Conjugate base + H3O+(e.g., sodium acetate + H3O+):A-+ H3O+HA + H2O K = 1/Ka= 1/(1 x 10-5) = 1 x 105(or bring to completion; then solve either using HH or reverse + Ka) Weak acid + strong base (e.g., acetic acid + sodium hydroxide): HA + OH-H2O + A-K = 1/Kcb= 1/(Kw/Ka) = Ka/Kw= (1 x 10-5)/(1 x 10-14) = 1 x 109(or bring to completion; then solve either using HH or reverse + Ka) Weak base + water (e.g., NH3+ H2O): RNH2+ H2O RNH3++ OH-Assume Kb= ~1 x 10-5Conjugate acid + water (e.g., sodium acetate + H2O): RNH3++ H2O RNH2+ H3O+Kca= Kw/Kb= (1 x 10-14)/(1 x 10-5) = (1 x 10-9) Conjugate acid + OH-(e.g., ammonium chloride + OH-): RNH3++ OH-RNH2+ H2O K = 1/Kb= 1/(1 x 10-5) = 1 x 105(or bring to completion; then solve either using HH or reverse + Kb) Weak base + strong acid (e.g., NH3+ H3O+): RNH2+ H3O+RNH3++ H2O K = 1/Kca= 1/(Kw/Kb) = Kb/Kw= (1 x 10-5)/(1 x 10-14) = 1 x 109(or bring to completion; then solve either using HH or reverse + Kb)
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Chem 162-2012 Chapter 16 (B&O) Version III 34Acid-base Properties of SaltsET: A salt is another name for an ionic compound, i.e., something that will dissociate into positive and negative ions in water. 111 (mod.) Are solutions of the following salts acidic, basic, or neutral? a. NaNO3NaNO3Na++ NO3-Na++ H2O NR NO3-+ H2O NR Neutral b. NaNO2NaNO2Na++ NO2-Na++ H2O NR NO2-+ H2O HNO2+ OH-Therefore, basic solution f. NH4ClO4NH4ClO4NH4++ ClO4-NH4++ H2O H3O++ NH3ClO4-+ H2O NR Therefore, acidic solution NH4CNNH4CN NH4++ CN- NH4++ H2O H3O++ NH3CN-+ H2O OH-+ HCN Note that in this case neither the acidic ion nor the basic ion drops out of the reaction. Which forms more, H3O+or OH-? NH4++ H2O H3O++ NH3Kb= 1.8x10-5; Kca = Kw/Kb= (1x10-14)/(1.8x10-5) = 5.56 x 10-10 CN-+ H2O OH-+ HCN Ka= 6.2x10-10; Kcb= Kw/Ka= (1x10-14)/(6.2 x 10-10) = 1.61x10-5Kcb> Kca. Therefore, more OH-is formed than H3O+. Therefore, NH4CN forms a basic solution.
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Chem 162-2012 Chapter 16 (B&O) Version III 35ET: Equilibria with salts; (1) dissolve in solution; (2) get rid of spectator ions; (3) write reaction. Use Kw. 105b. Calculate the pH of 0.050 M NaCN. K a for HCN = 6.2 x 10 -10 . NaCN Na + + CN - Na + + H 2 O NR CN - + H 2 O HCN + OH - CN - + H 2 O HCN + OH - Initial 0.050 Change Equilibrium CN - + H 2 O HCN + OH - Initial 0.050 0 0 Change -X +X +X Equilibrium 0.050-X +X +X K cb = ([HCN][OH - ])/[CN - ] What is the value of K cb ? K w = K a x K cb K cb = (1 x (10 -14 ))/(6.2 x (10 -10 )) K cb = 1.61 x 10 -5 (X x X)/(0.050-X) = 1.61 x 10 -5 Simplify with small K rule: (X x X)/0.050 = 1.61 x 10 -5 X = 8.89 x 10 -4 = [OH - ] [H + ] = (1 x 10 -14 )/(8.89 x 10 -4 ) = 1.12 x 10 -11 pH = -log(1.12 x 10 -11 ) = 10.95 Note: This is consistent with CN - being a base.
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Chem 162-2012 Chapter 16 (B&O) Version III 36 Given M and pH; find Ka. A 0.25M sodium hypochlorite solution (NaClO) has a pH of 10.43. Calculate K a for HClO (hypochlorous acid). NaOCl Na + + OCl - Na + = junk ion NaOCl Na + + OCl - OCl - + H 2 O HOCl + OH - pH = 10.43 [H + ] = 10 -pH = 10 -10.43 = 3.72 x 10 -11 [H + ][OH - ] = 1x10 -14 [OH - ] = (1x10 -14 )/[H + ] = (1x10 -14 )/(3.72 x 10 -11 ) = 2.69 x 10 -4 OCl - + H 2 O HOCl + OH - Initial 0.25 Change Equilibrium 2.69 x 10 -4 OCl - + H 2 O HOCl + OH - Initial 0.25 0 0 Change -X +X +X Equilibrium 0.25-X +X 2.69 x 10 -4 0 + X = 2.69 x 10 -4 ; X = 2.69 x 10 -4 OCl - + H 2 O HOCl + OH - Initial 0.25 0 0 Change -2.69 x 10 -4 2.69 x 10 -4 2.69 x 10 -4 Equilibrium 0.25-2.69 x 10 -4 = 0.25 2.69 x 10 -4 2.69 x 10 -4
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