Acids and Bases. H + (aq) + Cl - (aq) 100 molecules HCl 100 H+ ions Cl- ions 100% HCl molecules dissociate in water.

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2 Acids nd Bses Acids: in generl cid is the substnce tht produces H ions when it is dissolved in wter. Acids cn be divided into two different clsses: Strong cid: is one tht completely dissocites into its ions, such s HCl HCl (q) H (q) Cl - (q) 100 molecules HCl 100 H ions 100 Cl- ions 100% HCl molecules dissocite in wter Wek cid is one tht dissocites incompletely (slightly) into its ions, such s CH COOH (cetic cid, vinegr) CH COOH (q) H (q) CH COO - (q) 100 molecules CH COOH 1 H ions 1 CHCOO - ions 1% CH COOH molecules dissocite in wter (99 molecules CH COOH remin) 2

3 Acids nd Bses Bses: in generl bse is the substnce tht produces OH - ions when it is dissolved in wter. Bses cn be divided into two different clsses: Strong bse: is one tht completely dissocites into its ions, such s NOH NOH (s) H 2 O (l) N (q) OH - (q) 100 molecules NOH 100 N ions 100 OH - ions 100% NOH molecules dissocite in wter Wek bse is one tht dissocites incompletely (slightly) into its ions, such s NH (mmoni). NH (q) H 2 O (q) NH 4 (q) OH - (q) 100 molecules NH 1 NH 4 ions 1 OH -- ions 1% NH molecules dissocite in wter (99 molecules NH remin)

4 Autoioniztion of Wter Wter cn ct s either n cid or bse: 4

5 K w = [H O [OH - [H O = 1.0 x 10-7 M [OH - = 1.0 x 10-7 M Therefore, K w = (1.0 x 10-7 )(1.0 x 10-7 ) = 1.0 x Clculte the concentrtion of H ions in 1. M HCl. HCl (q) H 2 O (l) H O (q) Cl - (q) Initil: 1. M 0 M Finl: 0 M 1. M [H O = 1. M 5

6 Clculte the concentrtion of OH - ions in 1. M HCl. K w = [H O [OH - = 1.0 x (1. M) x [OH - = 1.0 x [OH - = (1.0 x / 1.) = 7.7 x M Clculte the concentrtion of OH - ions in 0.62 M NOH NOH (q) H 2 O (l) N (q) OH - (q) Initil: 0.62 M 0 M Finl: 0 M 0.62 M [OH - = 0.62 M Clculte the concentrtion of H ions in 0.62 M NOH K w = [H O [OH - = 1.0 x [H O x 0.62 M = 1.0 x [H = (1.0 x / 0.62) = 1.6 x M 6

7 ph nd poh [H O [OH - = K w = 1.0 x ph poh = ph = -log [H O poh = - log[oh - ph = -log [H O poh = - log[oh - 7

8 In neutrl solution: [H O = [OH - = 1.0 x 10-7 M Therefore, ph = poh = 7 In cidic solution: [H O > [OH - Therefore: [H O > 1.0 x 10-7 M ph < 7 [OH- < 1.0 x 10-7 M poh >7 In bsic solution: [H O < [OH - Therefore: [H O < 1.0 x 10-7 M ph > 7 [OH- > 1.0 x 10-7 M poh <7 8

9 If the concentrtion of H O of cn of coke is 7.6 x 10-4 M, is it cidic, bsic or neutrl? Clculte the ph nd the poh of 0.10 M solution of NOH in wter. 9

10 Ioniztion of wek cid nd bse Ioniztion of wek cids nd bses is not complete, some un-ionized cid remins in solution. CHCOOH( q) H2O HO ( q) CHCOO wekcid O [ ( q) [ K H [ C = C C H O H O K is Acetic cid ioniztion Cconstnt H = 1.8 x O O K b = K b is mmoni (bse) ioniztion constnt = 1.8 x [ NH 4 [ NH [ OH

11 Clculting ph nd poh for Wek Acids nd Bses 1. Write out the relevnt rections. cids donte proton to wter bses ccept proton from wter 2. Keep in mind tht uto-ioniztion of wter is lwys occurring to generte both H nd OH -. In most cses, these contributions will be negligible reltive to the wek cid or wek bse contributions, so we ignore them.. Use concentrtions tble to summrize initil nd equilibrium concentrtions of H or OH Use the equilibrium expression to clculte unknown vlues. 5. From the equilibrium concentrtions of H or OH -, clculte the ph or poh. 6. Use eqution ph poh = 14 to clculte the vlue (ph or poh) tht is still unknown. 7. Check your nswer: ph < 7 for cid; ph > 7 for bse! 11

12 Clculte the ph of 1.0 M solution of cetic cid in wter. Step 1: Determine the relevnt chemicl eqution(s) nd find the pk vlues CH COOH q) H O H O ( q) CH COO ( q) wekcid ( 2 K = 1.8 x 10-5 Step 2: Identify ll species present in the solution tht my ffect the ph. The concentrtion of H O is resulting from ioniztion of cetic cid. CH COOH q) H O( l) H O ( q) CH COO ( q) wekcid ( 2 The concentrtion of H O is resulting from ioniztion of cetic cid. Step : Summrize the chnges in concentrtions using the ICE tble. Where, x is concentrtion of cetic cid tht is consumed by the ioniztion of cetic cid in wter. 12

13 Step 4: Use the equilibrium expression to clculte the vlue for x. Step 5: Apply the pproximtion when possible. Mke the pproximtion tht x = 1.0 Step 6: Solve the eqution for x. Check your nswer: [H O =x = 4.2 x 10 - M Substitute vluefor x : Step 7: Determine ph. ph = - log [H O = - log (4.2 x 10 - ) = 2.4 (less thn 7.00, its cidic) x = 4.2 x 10-2 ( ) ( ) = correct! 1

14 Another esy nd relible solution for such problem is by using the pproximtion tht the initil concentrtion of either the wek cid or bse is very big compred with the chnge x. 2 X K = 2 X = K [wek cid [Wek cid [ H = X = K [wek cid So, using the bove eqution, the ph of 1.0 M solution of cetic cid in wter, will be: [H = 5 K [wek cid = 1.8 x10 x1.0 = 4.2 x10 ph = - log [H O = - log (4.2 x 10 - ) =

15 Clculte the ph of M solution of mmoni in wter. NH ( q) H 2O( l) NH 4 ( q) OH ( q) [ OH = Kb[wek bse So, using the bove eqution, the ph of 1.0 M solution of cetic cid in wter, will be: [OH = 5 Kb[wek bse = 1.8 x10 x 0.01 = 4.2 x10 4 poh = - log [OH - = - log (4.2 x 10-4 ) =.7 ph = poh = =

16 Henderson-Hsselblch Eqution For cetic cid (wek cid) CH COOH( q) H 2O H O ( q) CH COO ( q) [ CH COO [ HO K = [ CH COOH log[ H log[ H or [ H K [ CH = COOH O [ CH COO [ CH log [ CH COOH log O = K COO O = log K ph = pk [ CH log [ CH log COO COOH [ CH COO [ CH COOH 16

17 Clculte the ph of solution contining 0.20 M cetic cid (K = 1.8 x 10-5 ) nd 0.0 M sodium cette. ph = pk log [ CH COO [ CH COOH ph = ( 10 ) log log ( 0.0M ) ( ) 0.20M ph= =

18 Multiple Equilibri 1- Consider the following rections: nd A bb K 1 = [ B [ A b bb A K = 2 [A [B b In this cse, K 2 = 1 K 1 = [ A [ B b 18

19 Multiple Equilibri 2- Consider the following rections: A bb If this rection is multiply by : A bb In this cse, K 1 = [ B [ A b b b [B [B 2 = = K 1 K = [A [A ( ) 19

20 Multiple Equilibri - Consider the following rections: nd A bb K 1 = [ B [ A b bb cc The overll rection will be: A cc K K 2 = = [ C [ B [ C [ A c b c In this cse, K b c [ B [ C = K1 x K 2 = x = b [ A [ B [ C [ A c So, K = K 1 x K 2 20

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