Rank the following solutions in order of increasing pH: 0.1 M HNO 2 - - PDF document

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Rank the following solutions in order of increasing pH: 0.1 M HNO2, 0.1 M HNO3, 0.1 M NaNO2, 0.1 M NaNO3, 0.1 M NaOH

• A. HNO2 < HNO3 < NaNO2 < NaNO3 < NaOH
• B. HNO3 < HNO2 < NaNO2 < NaNO3 < NaOH
• C. HNO2 < HNO3 < NaNO3 < NaNO2 < NaOH
• D. HNO3 < HNO2 < NaNO3 < NaNO2 < NaOH
• E. HNO3 < HNO2 < NaOH < NaNO3 < NaNO2

2 4 6 8 10 12 14 10 20 30 40 50 60 70 80 90 100

pH Volume of 0.10 M NaOH added (mL) HCl HNO2

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 10 20 30 40 50 60 70 80 90 100 110

pH volume of 0.1 M NaOH(aq) added

2.19 48 4.73 49 5.04 49.9 6.05 49.99 7.04 49.999 7.82 50 8.01 50.001 8.20 50.01 8.97 50.1 9.97 51 10.96 52 11.26 65 12.08 70 12.18 75 12.26 100 12.48 1000 12.90 10000 12.94

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You can calculate the pH of 0.10 M NH4Cl(aq) using the equation given on the right where K and x are A. Ka [H+] B. Ka [OH] C. Kb [H+] D. Kb [OH]

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0.10 x K x  

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m = 1 2 (c1z1

2 + c2z2 2 +×××)  1

2 å

i cizi 2

Ionic strength (m) is given by the equation where ci is the concentration of ion i, and zi is the charge of ion i.

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To determine if 0.10 M HCO3

 is acid or basic,

you must compare Ka2 for HCO3

 against the Kb

for HCO3

 which is given by

• A. Kb = Kw/Ka1
• B. Kb = Kw/Ka2
• C. Kb = Ka1/Kw
• D. Kb = Ka2/Kw

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H2CO3(aq) + OH−(aq) −−> HCO3

−(aq) + OH−(aq) −−> CO3 2−(aq)

+ H2O(l) + H2O(l) H2SO3(aq) + OH−(aq) −−> HSO3

−(aq) + OH−(aq) −−> SO3 2−(aq)

+ H2O(l) + H2O(l)

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H2SO3(aq) + OH−(aq) −−> HSO3

−(aq) + OH−(aq) −−> SO3 2−(aq)

+ H2O(l) + H2O(l)

Recall…

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Ka1 = 1.4  10 2 Ka2 = 6.5  10 8 H2SO3 HSO3



SO3

2

H2A HA A2 Ka1 = 1.0  10 4 Ka2 = 1.0  10 8

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H2A HA A2 Ka1 = 1.0  10 5 Ka2 = 1.0  10 7 Ka1 = 1.0  10 6 Ka2 = 1.0  10 6

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m = 1 2 (c1z1

2 + c2z2 2 +×××)  1

2 å

i cizi 2

Ionic strength (m) is given by the equation where ci is the concentration of ion i, and zi is the charge of ion i.

H2SO3(aq) + OH−(aq) −−> HSO3

−(aq) + OH−(aq) −−> SO3 2−(aq)

+ H2O(l) + H2O(l)

Recall…

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Ka1 = 1.4  10 2 Ka2 = 6.5  10 8 H2SO3 HSO3



SO3

2

H2A HA A2 Ka1 = 1.0  10 4 Ka2 = 1.0  10 8

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H2A HA A2 Ka1 = 1.0  10 5 Ka2 = 1.0  10 7 Ka1 = 1.0  10 6 Ka2 = 1.0  10 6

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copper wire silver wire 1 M Cu(NO3)2 1 M AgNO3 Cu2+ ions → ← NO3

• ions

copper metal is oxidized at the anode: Cu(s) → Cu2+(aq) + 2e‒ silver ions are reduced at the cathode: Ag+(aq) + e‒ → Ag(s) 0.46 V

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0 4 6

Cu(s) 1 M Cu2+ 1 M Ag+ Ag(s)

1 5 6

Zn(s) 1 M Zn2+ 1 M Ag+ Ag(s)

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Ag+(aq) + 1e  Ag(s) Cu2+(aq) + 2e  Cu(s) Zn2+(aq) + 2e  Zn(s) 0.46 V 1.56 V Ag+(aq) + 1e  Ag(s) Cu2+(aq) + 2e  Cu(s) Zn2+(aq) + 2e  Zn(s) 0.46 V 1.56 V 1.10 V

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1 1 0

Zn(s) 1 M Zn2+ 1 M Cu2+ Cu(s) Ag+(aq) + 1e  Ag(s) Cu2+(aq) + 2e  Cu(s) 2H+(aq) + 2e  H2(g) Zn2+(aq) + 2e  Zn(s) 0.36 V 1.56 V 1.10 V

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Ag+(aq) + 1e  Ag(s) Cu2+(aq) + 2e  Cu(s) 2H+(aq) + 2e  H2(g) 0.000 V Zn2+(aq) + 2e  Zn(s) 0.36 V 1.56 V 1.10 V Ag+(aq) + 1e  Ag(s) 0.80 V Cu2+(aq) + 2e  Cu(s) 0.34 V 2H+(aq) + 2e  H2(g) 0.000 V Zn2+(aq) + 2e  Zn(s) 0.76 V 0.46 V 1.56 V 1.10 V

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Ag+(aq) + 1e  Ag(s) 0.80 Cu2+(aq) + 2e  Cu(s) 0.34 2H+(aq) + 2e  H2(g) 0.000 Zn2+(aq) + 2e  Zn(s) 0.76 E°(V)

22 Decreasing potential  wants electrons less  weaker oxidizing agent  product is a stronger reducing agent



Increasing potential  wants electrons more  stronger oxidizing agent  Decreasing potential  wants electrons less  weaker oxidizing agent  product is a stronger reducing agent



Increasing potential  wants electrons more  stronger oxidizing agent 

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