[base (aq)][oh (aq)] [base(aq)] [OH ( aq )]
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1 Section 8.5: Clcultions Involving Bsic Solutions Tutoril 1 Prctice, pge Given: [OH (q)] mol/l Required: [H + (q)], [OH (q)] [OH (q)] mol/l ecuse OH is strong se [H ] [OH ( q )] Solution: [H ] [H + (q)] mol/l [OH (q)] 1.00 mol/l 2. Given: [Sr(OH) 2 (q)] mol/2.00l mol/l [OH (q)] mol/l ecuse Sr(OH) 2 is strong se producing 2 mol OH per mole of compound Solution: poh log[oh (q)] log(0.042) poh 1.77 ph poh ph Tutoril 2 Prctice, pge () From tle, (C 2 H O 2 ) 5.6 () From tle, (H BO ) 5.8 w Given: [se] 0.20 mol/l;.82 se(q) #!!! "! se + (q) + OH (q) I C x + x + x E 0.20 x x x + [se (q)][oh (q)] [se(q)] Copyright 2012 Nelson Eduction Ltd. Chpter 8: Acid-Bse Equilirium 8.5-1
2 Solution: ( x)( x).82 (0.20 x) ()() x x x x [OH (q)] mol/l poh log( ) poh 4.06 ph ph Given: [N 2 H 4 ] 4.5 mol/l; N 2 H 4 (q) + H 2 O(l) #!!! "! N 2 H + 5 (q) + OH (q) I C x + x + x E 4.5 x x x [N2H5 [N2H4 q ] 6 ( x)( x) Solution: 1.7 (4.5 x) 6 ( x)( x) x x [OH (q)] 2.77 mol/l poh log(2.77 ) poh 2.56 ph ph Copyright 2012 Nelson Eduction Ltd. Chpter 8: Acid-Bse Equilirium 8.5-2
3 Section 8.5 Questions, pge () Given: [NOH (q)] mol/l [OH (q)] mol/l ecuse OH is strong se. Solution: poh log[oh (q)] log(0.0000) 2.52 ph poh ph () Given: [B(OH) 2 (q)] mol/l [OH (q)] mol/l ecuse B(OH) 2 is strong se producing 2 mol OH per mole of compound Solution: poh log[oh (q)] log(0.0040) 2.40 poh 2.40 ph poh ph (c) Given: [CH NH 2 (q)] 0.0 mol/l; CH NH 2 (q) #!!! "! CH NH + (q) + OH (q) I C x + x + x E 0.0 x x x [CHNH [CHNH 2 q ] 4 ( x)( x) Solution: 9.6 (0.0 x) 4 ( x)( x) x x [OH (q)]. mol/l poh log(. ) poh 2.51 ph ph Copyright 2012 Nelson Eduction Ltd. Chpter 8: Acid-Bse Equilirium 8.5-
4 (d) Given: [N 2 H 4 ] mol/l; N 2 H 4 (q) + H 2 O(l) #!!! "! N 2 H + 5 (q) + OH (q) I C x + x + x E x x x [N2H5 [N2H4 q ] 6 ( x)( x) Solution: 1.7 ( x) 6 ()() x x x x [OH (q)] mol/l poh log( ) poh.69 ph ph.1 2. () From tle, (H 2 S) w () w (c) From tle, (HCN) 6.2 w Copyright 2012 Nelson Eduction Ltd. Chpter 8: Acid-Bse Equilirium 8.5-4
5 (d) From tle, (HF) w () Given: [C 17 H 19 NO (q)] 0.01 mol/l; C 17 H 19 NO (q) + H 2 O(l) #!!! "! C 17 H 20 NO + (q) + OH (q) I C x + x + x E 0.01 x x x [C17H20NO [C17H19NO q ] 7 ( x)( x) Solution: 7.5 (0.01 x) 7 ( x)( x) x x [OH (q)] mol/l poh log(8.7 5 ) poh 4.06 ph ph 9.94 () Given: [C 21 H 22 N 2 O 2 (q)] mol/l; C 21 H 22 N 2 O 2 (q) + H 2 O(l) #!!! "! C 21 H 2 N 2 O + 2 (q) + OH (q) I C x + x + x E x x x [C21H2N2O2 [C H N O q ] Copyright 2012 Nelson Eduction Ltd. Chpter 8: Acid-Bse Equilirium 8.5-5
6 Solution: 6 ()() x x 1.0 (0.001 x) 6 ( x)( x) x x [OH (q)].2 5 mol/l poh log(.2 5 ) poh 4.50 ph ph () Given: ([C 2 H 5 ) N(q)] 0.20 mol/l; Required: [H + (q)], [OH (q)], ph (C 2 H 5 ) N(q) + H 2 O(l) #!!! "! (C 2 H 5 ) NH + (q) + OH (q) I C x + x + x E 0.20 x x x ( 2 5) [CH [ C H NH N q] Solution: ()() x x 4.0 (0.20 x) 4 ( x)( x) x x [OH (q)] 8.94 mol/l [H + (q)] [OH (q)] [H (q)] 8.94 [H + (q)] mol/l ph log( ) Sttement: [OH (q)] 8.9 mol/l; [H + (q)] mol/l; ph Copyright 2012 Nelson Eduction Ltd. Chpter 8: Acid-Bse Equilirium 8.5-6
7 () Given: [HONH 2 (q)] 0.20 mol/l; Required: [H + (q)], [OH (q)], ph HONH 2 (q) #!!! "! NH + 2 (q) + OH (q) I C x + x + x E 0.20 x x x [NH2 [HONH2 ( q )] 8 ()() x x Solution: 1.1 (0.20 x) 8 ( x)( x) x x [OH (q)] mol/l [H + (q)] [OH (q)] [H (q)] [H + (q)] 2.1 mol/l ph log(2.1 ) 9.67 Sttement: [OH (q)] mol/l; [H + (q)] 2.1 mol/l; ph Given: [C(OH) 2 ] mol/l; Required: [OH (q)], poh, ph Solution: [OH (q)] mol/l ecuse C(OH) 2 is strong se. poh log( ). ph ph.90 Sttement: [OH (q)] mol/l; poh.; ph.90 Copyright 2012 Nelson Eduction Ltd. Chpter 8: Acid-Bse Equilirium 8.5-7
8 6. () Given: [OH] 25 g/l Required: [OH (q)], poh, ph 25 g 1 mol Solution: 0.45 mol/l L 56.0 g [OH (q)] 0.45 mol/l ecuse OH is strong se. poh log(0.45) 0.5 ph ph 1.65 Sttement: [OH (q)] 0.45 mol/l; poh 0.5; ph 1.65 () Given: [NOH] g/l Required: [OH (q)], poh, ph g 1 mol Solution:.75 mol/l L g [OH (q)].75 mol/l ecuse NOH is strong se. poh log(.75) 0.57 ph (0.57) ph Sttement: [OH (q)].75 mol/l; poh 0.57; ph Given: ph 11.80; Required: [NH (q)] NH (q) + H 2 O(l) NH + 4 (q) + OH (q) Solution: poh [OH (q)] 2.20 [OH (q)] 6. mol/l [NH4 ( q )][OH (q)] [NH q ] (6. )(6. ) [NH q ] (6. )(6. ) 5 [NH q ] 1.8 [NH ] 2.2 mol/l 8. Answers will vry. Smple nswer: Use chemicl sfety goggles nd/or full fce shield where splshing is possile. Wer protective clothing, including gloves, l cot, or pron. Work in fume hood. Copyright 2012 Nelson Eduction Ltd. Chpter 8: Acid-Bse Equilirium 8.5-8
9 9. () () (c) 2 + ( q )][H (q)] 2 2 [HCO2H [HCO q ] [HCO2 [HCO H q ] + [HCO H [HCO2 q ][H (q)] [HCO q ] [HCO H q ] + [OH (q)] [H (q)] w () Given: ph 11.2; Required: [CH NH 2 (q)] CH NH 2 (q) + H 2 O(l) CH NH + (q) + OH (q) Solution: poh [OH (q)] 2.80 [OH (q)] 1.6 mol/l 4.4 [NH 4 [NH q ] 4 (1.6 )(1.6 ) 2 [CH NH q ] (1.6 )(1.6 ) 2 4 [CH NH q ] 4.4 [CH NH 2 ] mol/l; this is not the correct concentrtion. () The solution is too dilute. (c) Answers my vry. Smple nswer: A Cost Gurd cocktil is comintion of drugs, including ephedrine tht is used to prevent sesickness. The ephedrine prevents the drowsiness tht is side effect of the other drug. It is not now generlly used ecuse ephedrine cn cuse nuse, insomni, confusion, nd occsionlly hypertension. 2 Copyright 2012 Nelson Eduction Ltd. Chpter 8: Acid-Bse Equilirium 8.5-9
30. a. Al(H 2 O) 6 + H 2 O Co(H 2 O) 5 (OH) 2+ + H 3 O + acid base conjugate base conjugate acid
Chpter 14 : xx,3,40,4,xx,xx,47,50,54,55,xx,64,xx,83,85,91,93,95,10,11,134,138 xx.. HC H 3 O (q) H O(l) C H 3 O (q) H 3 O (q) = [H O ][C H O ] 3 3 [HC H O ] 3 b. Co(H O) 6 3 (q) H O(l) Co(H O) 5 (OH) (q)
Which of the following describes the net ionic reaction for the hydrolysis. Which of the following salts will produce a solution with the highest ph?
95. Which of the following descries the net ionic rection for the hydrolysis of NH4Cl( s)? A. NH4 ( q) Cl & ( q) NH4Cl( s) B. NH Cl & 4 ( s) NH4 ( q) Cl ( q) C. Cl ( q) H O & 2 ( l) HCl( q) OH ( q) D.
HONH 3 initial M - ~0 0 -x - +x +x equilibrium x - x x. ] (x)(x) x b. If you assume that x << 0.100, then x
![HONH 3 initial M - ~0 0 -x - +x +x equilibrium x - x x. ] (x)(x) x b. If you assume that x << 0.100, then x](/thumbs/72/67573254.jpg "HONH 3 initial M - ~0 0 -x - +x +x equilibrium x - x x. ] (x)(x) x b. If you assume that x << 0.100, then x")
Chpter 15 : xx, 4, 8, 0, 8, 44, 48, 50, 51, 55, 76, 8, 90, 96(90), 1(111) xx. NH 4 OH NH H O NH H O NH 4 H O 4.. HONH (q) H O(l) OH (q) HONH (q) HONH H O OH HONH initil 0.100 M ~0 0 x x x equilibrium 0.100
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[H3 O (aq)] = = = mol/l. = 0.13 mol mol [OH (aq)] = [NaOH(aq)] = = mol/l 4.00 L 14
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