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

Rank the following solutions in order of increasing pH: 0.1 M HNO 2 , 0.1 M HNO 3 , 0.1 M NaNO 2 , 0.1 M NaNO 3 , 0.1 M NaOH A. HNO 2 < HNO 3 < NaNO 2 < NaNO 3 < NaOH B. HNO 3 < HNO 2 < NaNO 2 < NaNO 3 < NaOH C. HNO 2 < HNO 3 < NaNO 3 < NaNO 2 < NaOH D. HNO 3 < HNO 2 < NaNO 3 < NaNO 2 < NaOH E. HNO 3 < HNO 2 < NaOH < NaNO 3 < NaNO 2 14 12 10 8 pH HCl 6 HNO2 4 2 0 0 10 20 30 40 50 60 70 80 90 100 Volume of 0.10 M NaOH added (mL) 1

14 48 4.73 13 49 5.04 12 49.9 6.05 65 12.08 11 49.99 7.04 70 12.18 10 49.999 7.82 75 12.26 9 50 8.01 100 12.48 8 pH 7 50.001 8.20 1000 12.90 6 50.01 8.97 10000 12.94 5 50.1 9.97 4 0 2.19 51 10.96 3 52 11.26 2 1 0 0 10 20 30 40 50 60 70 80 90 100 110 volume of 0.1 M NaOH(aq) added 2

You can calculate the pH of 0.10 M NH 4 Cl( aq ) using 2 x the equation given on the  K  right where K and x are 0.10 x A. K a [H + ] [OH  ] B. K a C. K b [H + ] [OH  ] D. K b 3

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m = 1 2 + c 2 z 2 2 +××× )  1 2 å 2 2 ( c 1 z 1 i c i z i Ionic strength ( m ) is given by the equation where c i is the concentration of ion i, and z i is the charge of ion i . 5

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 is acid or basic, To determine if 0.10 M HCO 3  against the K b you must compare K a2 for HCO 3  which is given by for HCO 3 A. K b = K w /K a1 B. K b = K w /K a2 C. K b = K a1 /K w D. K b = K a2 /K w 7

H 2 CO 3 (aq) + OH − (aq) −−> HCO 3 − (aq) + OH − (aq) −−> CO 3 2− (aq) + H 2 O( l ) + H 2 O( l ) H 2 SO 3 (aq) + OH − (aq) −−> HSO 3 − (aq) + OH − (aq) −−> SO 3 2− (aq) + H 2 O( l ) + H 2 O( l ) 8

Recall… H 2 SO 3 (aq) + OH − (aq) −−> HSO 3 − (aq) + OH − (aq) −−> SO 3 2− (aq) + H 2 O( l ) + H 2 O( l ) 9

K a1 = 1.4  10  2 K a2 = 6.5  10  8 H 2 SO 3 HSO 3  SO 3 2  K a1 = 1.0  10  4 K a2 = 1.0  10  8 H 2 A HA  A 2  10

K a1 = 1.0  10  5 K a2 = 1.0  10  7 H 2 A HA  A 2  K a1 = 1.0  10  6 K a2 = 1.0  10  6 11

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m = 1 2 + c 2 z 2 2 +××× )  1 2 å 2 2 ( c 1 z 1 i c i z i Ionic strength ( m ) is given by the equation where c i is the concentration of ion i, and z i is the charge of ion i . Recall… H 2 SO 3 (aq) + OH − (aq) −−> HSO 3 − (aq) + OH − (aq) −−> SO 3 2− (aq) + H 2 O( l ) + H 2 O( l ) 13

K a1 = 1.4  10  2 K a2 = 6.5  10  8 H 2 SO 3 HSO 3  SO 3 2  K a1 = 1.0  10  4 K a2 = 1.0  10  8 H 2 A HA  A 2  14

K a1 = 1.0  10  5 K a2 = 1.0  10  7 H 2 A HA  A 2  K a1 = 1.0  10  6 K a2 = 1.0  10  6 15

0.46 V Cu 2+ ions → - ions ← NO 3 copper wire silver wire 1 M AgNO 3 1 M Cu(NO 3 ) 2 copper metal is oxidized at the anode: silver ions are reduced at the cathode: Cu(s) → Cu 2+ (aq) + 2e ‒ Ag + (aq) + e ‒ → Ag(s) 16

Cu(s) Ag(s) 0 4 6 1 M Cu 2+ 1 M Ag + Zn(s) Ag(s) 1 5 6 1 M Zn 2+ 1 M Ag + 17

Ag + ( aq ) + 1 e   Ag( s ) 0.46 V Cu 2+ ( aq ) + 2 e   Cu( s ) 1.56 V Zn 2+ ( aq ) + 2 e   Zn( s ) Ag + ( aq ) + 1 e   Ag( s ) 0.46 V Cu 2+ ( aq ) + 2 e   Cu( s ) 1.56 V 1.10 V Zn 2+ ( aq ) + 2 e   Zn( s ) 18

Zn(s) Cu(s) 1 1 0 1 M Zn 2+ 1 M Cu 2+ Ag + ( aq ) + 1 e   Ag( s ) 0.36 V Cu 2+ ( aq ) + 2 e   Cu( s ) 1.56 V 2H + ( aq ) + 2 e   H 2 ( g ) 1.10 V Zn 2+ ( aq ) + 2 e   Zn( s ) 19

Ag + ( aq ) + 1 e   Ag( s ) 0.36 V Cu 2+ ( aq ) + 2 e   Cu( s ) 1.56 V 2H + ( aq ) + 2 e   H 2 ( g ) 0.000 V 1.10 V Zn 2+ ( aq ) + 2 e   Zn( s ) Ag + ( aq ) + 1 e   Ag( s ) 0.80 V 0.46 V Cu 2+ ( aq ) + 2 e   Cu( s ) 0.34 V 1.56 V 2H + ( aq ) + 2 e   H 2 ( g ) 0.000 V 1.10 V Zn 2+ ( aq ) + 2 e   Zn( s )  0.76 V 20

E°(V) Ag + ( aq ) + 1 e   Ag( s ) 0.80 Cu 2+ ( aq ) + 2 e   Cu( s ) 0.34 2H + ( aq ) + 2 e   H 2 ( g ) 0.000 Zn 2+ ( aq ) + 2 e   Zn( s )  0.76 21

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

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