Chemical equilibrium and Acids, bases, Exercises of Chemical Principles

Download Chemical equilibrium and Acids, bases and more Exercises Chemical Principles in PDF only on Docsity! The following concentrations were obtained for the eee of NH, from N, and ia at equilibrium at S00K, [N,]=1.5x10?M.[H.,] = 3.0 x 10° M and [NH,] = 1.2 x 102m, Calculate equilibrium constant. The equilibrium constant for the reaction, Nj... + 2H, ,<—— 2NH,,,, can be written as, = INH, __(1.2«107)"___ = 0.031810" = 3.1810? “"(N,][H,] (1.5x107)(3.0x107) At equilibrium, the concentrations of N, = 3.0 x 10°M, O, = 4.2 x 10°M and NO = 2.8 x 10°M in a sealed vessel at 800K. What will be K, forthe reaction. —_N,,. + O,,, <=> 2NO,, ; Sag. 1) ) For the reaction equilibrium constant, K, can be written as, « et. ery * [N,][0,] (3.0x107)(4.2x107) | pemattrmmhrs Manian "J 0.622 PCI,, PC/, and Ci, are at equilibrium at 500K and having concentration 1.59M PCI, 1.59M Ci, and 1.41 M PC/i,. Calculate Ke for the reaction. PC/, == PC/, + Ci, The equilibrium constant K, for the PC/, —— = PCI, + Ci, can be written as, K = [PCI,}[CZ, ] a (1.59)(1.59) “3 (PCI, ] d.4)) eegngne EnyEE 1.79 M. The value of K, = 4.24 at 800K for the reaction, Co,,, . H,0,, = =C0,, + H Calculate equilibrium concentrations of CO,, H,, CO and H,® at 800K K, it only CO and H,0 are present initially at concentrations of 0.10M each. For the reaction, co, . + H,O, 2 <== 60F i + Hy Initial concentration: 0.1M 0.1M 0 0 Let x mole per litre of each of the product be formed. a age ; (CO, ]{H,] At lib: ilibrium constant can be written as, K, = ——2——— equilibrium, equ [CO][H,0] 4.24=x?/(0.1-x)? = x? =4.24(0.01+ x? -0.2x) => x? =0.0424 +4.2x? -0.848x => 3.24x? -0.0848x +0.0424 =0 a= 3.24, b =-0.848, c = 0.0424 for quadratic equation (ax? + bx +c =0), x=(-b+ Vb? = 4ac) /2a x =0.848+ (0.848)? — 4(3.24)(0.0424) /(3.24x 2) => x = (0.848 + 0.4118) / 6.48 x, = (0.848 -- 0.4118) / 6.48 = 0.067 — x, = (0.848 + 0.4118) / 6.48 = 0.194 the value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration. Hence the equilibrium concentrations are, [CO,]=[H,]=x = 0.067M [CO] = [HO] = 0.1 — 0.067 = 0.033M 7 A. ee a Kor the equilibrium 2NOC/,, <== 2NO,, + C/,,) the value of the equilibrium constant, K, is 3.75 x 10° at 1069K. Calculate the K, for the reaction at this temperature? We know that, K, = K, (RT) for the above reaction, An = (2 + I-2=1 K, = 3.75 x 10+ (0.0831 x 1069) K, = 3.3 x 10“ 6. ape 6, == 2C0,, is 3.0 at 1000K. If initially 0 " Poo, = 0.48 bar and P., =6 bar and pure graphite is present, calculate the equilibrium Partial pressures of CO and co,. For the reaction, let ‘x’ be the decrease in pressure of CO,, then co, +C, == 200, Xe) The value of K, for the reaction, CO. initial pressure : 0.48 bar 0 At equilibrium : (0.48 — x) bar 2x bar 2 K, = PE —53=(2%)9(0.48 ~ x) = 4x°=3(0.48- x) => 4x? =1.44 —x =p 4x43x-1.4d= 0 co; a=4, b=3, c=-1.44 for quadratic equation (ax? + bx + ¢ = 0), x=(-b+Vb? —4ac)/2a x=[-34/@) -4(4)(-1.44))/2x4 => x =(-345.66)/8 as value of ‘x’ cannot be negative, x = 2.66 / 8 = 0.33 The equilibrium partial pressures are, Poo = 2x =2x 0.33 = 0.66 bar Pco, = 0.48-x = 0.48 -0.33= 0.15 bar. 7. The value of K, for the reaction 2A == B + C is 2 x 10°. At a given time, the composition of i caction mixture is [A] = [B] =[C] =3x 10M. In which direction the reaction will pro. ~cd? A. For the reactino the reaction quotient Q, is given by, Q. = [B][C] / [A as [A] = {B] = [C] =3 x 10“M Q, = (3 x 10) (3 x 10) /(3 x 104)? = 1 Q. > K, so the reaction will proceed in the reverse direction. 8. 13.8g o N,O, was placed in a 1L reaction vessel at 400K and allowed to ddsaitn equilibrium N,O, == 2NO, ,). The total pressure at equilibrium was found to be 9.15 bar. Calculate K., K, and partial pressure at equilibrium. A. Total volume (V) = 1 L, Molecular mass of N,O, = 92g ; Temperature (T) = 400K No. of moles (n) = 13.8g/ 92g = 0.15; Gas constant (R) = 0.083 bar L mot! ee We know that, PV = nRT = Px 1=0.15 x 0.083 x 400 = P= 4.98 bar N,O, == 2NO, Initial pressure : 4.98 bar 0 At equilibrium : (4.98 ~ x) bar 2x bar HF+ HO == HO + F (HO }[F Initial concentration: © (M) 0.02 0 cf ad {HF} Change : (M) -0.02¢@ +0.02a +0.02a Equilibrium concentration(M) 0.02 -0.02a 0.02a 0.02a Substituting equilibrium concentrations in the equilibrium reaction for principal reaction gives K, = (0.02)? / (0.02 - 0.02) = 0.02 a? / (1 - a) = 3.2 x 10“ We obtain the following quadratic equation : a? + 1.6 x 10°a - 1.6 x 10°=0 The quadratic equation in a can be solved and the two values of the roots are : a = +0.12 and - 0.12 The negative root is not acceptable and hence, a = 0.12 It means that the degree of ionization, a = 0.12, then equilibrium concentrations of other species viz., HF, F and H,O* are given by : (H,O*} = [F] = ca = 0.02 x 0.12 = 2.4 x10°M [HF] = c(1 - aw) = 0.02(1 - 0.12) = 17.6 x 10°M pH = — tog[H*] = — log(2.4 x 10") = 2.62 19. The pH of 0.1M monobasic acid is 4.50. Calculate the concentration of specics H*, A- and HA at equilibrium. Also, determine the value of K, and pK, of the monobasic acid. pH = - log[H’), Therefore, [H*] = 10 t= 10+ = 3.16 x 10° [H*] = [A] = 3.16 x 10° Thus, K, = [H*] [A)] / [HA] [HA],,,,, = 0.1 - (3.16 x 10°) = 0.1 K, = (3.16 x 10° /0.1=1.0x 10% — -. pK, =~ log(10“) = 8° Alternatively, “Percent dissociation” is another useful method for measure of strength of a weak acid and is given as : Percent dissociation = [HA], .osiaeq * 100 / [HA] ig (7-32) Calculate the pH of 0.08M solution of hypochlorous acid, HOCL The sateen constant of the acid is 2.5 x 10°. Determine the percent dissociation of HOCL HOC? (aq) + H,0 (/) ==? H,O'(aq) + ClO (aq) Initial concentration (M) 0.08 0 0 0 Change to reach equilibrium conc, (M) x x x Xx equilibrium concentration (M) 0.08 — x x x x Ka = {[H,O°] [ClO™) / [HOCI]} = x? / (0.08 — x) As x << 0.08, therefore 0.08 = 0,08 x7 /0.08=2.5« 10'=> x°=2.0 x 10% thus, x = 1.41 x 10° => [H*] = 1.41 x 10°'M. ~. percent dissociation={{HOCI] ,_....4% 100/{HOCI,}= L4l x 10°/0,08= 1.76%. pH =—log(1.41 x 10°) = 2.85 21s Tpsplt of 0.004M hydrazine solution is 9.7. Calculate its ionization constant K, and “ph,. 22. Calculate the pH of the solution in which 0.2M NH,Cl and 0.1M NH, are present. A. NH,+H,O == NH, + OH- a's VileiiiCal CQUINDMUM & ACIGS Sas 23. Determine the cegreed of ionization and pH of a 0.05M of ammonia solution. The A. The ionization of NH, in water is represented by equation : NH, + HO—= NH,’ + OH &. NH.NH, + H,O <== NHNH,‘ + OH : From the PH we can calculate the hydrogen ion concentration. Knowing a ion Concentration and the ionic product of water we can calculate the concentration of hydroxyl lons Thus we have ; [H*] = antilog (-pH) = antilog (-9.7) = 1.67 x 10" (OH ] = K, /[H*] = 1 x 10" / 1.67 x 10- = 5.98 x 10° The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentrations of both these ions are very small so the concentration of the undissociated base can be taken equal to 0.004M. Thus, K, = [NH, NH,*] [OH-] / [NH,NH,] = (5.98 x 10-5)? / 0.004 = 8.96 x 107 PK, = - log K, = — log(8.96 x 10-7) = 6.04. The pK, of ammonia solution is 4.75. The ionization constant of NH,, K, = antilog (-pK,) ie., K, = 1047 = 1.77 x 10°M NH, + H,O—= NH, + OH Initial concentration (M) 0.10 0.20 0 0 Change in conc. to reach equilibrium (M) -x +x +x x At quilibrium (M) 0.10-x 0.20-x +x x K, = [NH,*] [OH] / [NH,] = (0.20 + x) (x) / (0.1 - x) = 1.77 x 10° As K, is small, we can neglect x in comparison to 0.1M and 0.2M. Thus, [OH] =x =0.88 x 10° ~. (H*] =1.12 x 10° pH = - log[H*] = 8.95. ionization constan: of ammonia 1.77 x 10°. Also, calculated the ionization constant of the conjugate acid of ammonia. hydroxyl ion concentration, [OH™] = ca = 0.05a The value of @ is calculated from the equation, K, =ca?/(1—a), The value of a i: small, therefore the quadratic equation can be simplified by neglecting a in comparison to | in the denominator on right hand side of the equation, Thus, K,=c a? = a= JKy/c => a= V1.77x10°/0.05 = 0.018. [OH] = c a = 0.05 x 0.018 = 9.4 x 104M. [H"] = K, / [OH™] = 107 (9.4x10") = 1.06x10" ». pH = - log (1.06x 10") = 10.97, Now, using the relation for conjugate acid - base pair, K,x K,=K_ using the value of K, of NH, from table 7.7. We can determine the concentration of conjugate acid NH, K,=K,/K, = 10"/ 1.77 x 10° = 5.64 x 10-1, — = a ; 24, Calculate the pH of a 0.10M ammonia solution. Calculate the pHi after 50.0m), oft 44 tion constany o¢ | this solution is treated with 25.0ml. of 0.10M HCL The dissocia ammonia, K, = 1.77x 10‘ A. NH, +H,O == NH,’ OH K, ENH"| [OT | (NH | 177 «10 Let [NH,‘]= [OH] =x © [NHJ 0.10. x = 0.10 K,=x7/0.10 1.77 10° 2/010 > x=133% 10 = [OH | (H*) =K, / (OH J} = 107 (1.33 « 10%) = 7.51 x 10°? *. pH =~ log(7.5 x 10°) = 11.12 On addition of 25mL of 0.1M HCI solution (i.e., 2.5 mmol of HCl) to 50 mL. of 0. IM ammonia solution (i.e., 5 mmol of NH,), 2.5 mmol of ammonia molecules are neutralized. The resulting 75 mL solution contains the remaining unneutralized 2.5mmol of NH, molecules and 2.5 mmol of NH,’. NH, + HCl > NEY + Cr 258 2S 0 0 The resulting 75mL of solution contains 2.5 m mol of NH,” ions (i.e., 0.033M) and 2.5 mmol (i.e., 0.033M) of unneutralized NH, molecules. This NH, exists as NH ,OH in the following equilibrium : NHOH = = NH/ + OH 0.0333M-y y y where, y = [OH] = [NH,"] The final 75 MI solution after neutralization already contains 2.5 mmol MH, ions (Le., 0.033M), thus total concentration of NH,’ ions is given as - [NH,‘] = 0.033 + y As y is small, [NH,OH] = 0.033M and [NH,'] ~ 0.033M. We know, K, [NH,"] [OH"] / [NH,OH] = y (0.033) / (0.033) = 177 « i0 3M Thus, y = 1.77 x 10% = [OH] [H*] = 10°/ 1.77 x 10° = 0.56 x 10° Hence, pH = 9.24 25. The pK, of acetic acid and PK, of ammonium hydroxide are 4.76 and 4.75 respectively. | Calculate the pH of ammonium acetate solution. | A. pH =7+ ¥4[pK, ~pK,] =7+ 414.76 -4.75] =74 ¥4[0.01] = 7 + 0.005 = 7.005 26. Calculate the solubility of A,X, in pure water, assuming that neither kind of ion | reacts with water. The solubility product of A,X, K. = Lx 107, A, AX, == 2A + 3x? K, = {A"? [X? P= 1b x 102 if S = solubility of A,X,, then [A*] = 2S; [X?] = 3S therefore, K, = (28/38) = 108 S°= 1.1 x 10% | thus S*= 1 x 10° S = 1.0 x 10° mol/L scien eemmemammiati 4 At certain temperature, K, for the reaction. $0,()+No 2(8) = SO,(g)+NO(g) {s 16. If initially one mole each of all the four gases -—~are taken in one litre vessel. What are the equilibrium concentrations of NO and SO,. ~ SOx) + NOx) === SOyg) + NOw) Initial moles : 1 1 1 1 Moles at equilibrium : (l-x) (1-x) a: ts ae l-x l-x (1+x I+x Equilibrium conc.(mol/lit) : (4) i: or! 1 Given, K, = 16 Equilibrium constant, K, - [SO, ][NO,] _(1+x+x) 42 _ (+x)? 1+x = 4% = > 4a (l-x)(1—x) x) (I-x)2 1-x 4-4x=1+x >5x=3 1x =3/5=0.6 Concentration of [NO] = 1 + 0.6 = 1.6 mole/lit Concentration [SO,] = 1 + 0.6 = 1.6 mole/lit 5. Under certain conditions, 1 the equilibrium constant for the decomposition of PC/ (2) into PCI, (g) and Cl, (g) is 0.0211 mol L-'. What are the equilibrium concentrations of PCi,, PCI, and C/. if the initial concentration of PC/, was 1.00 M? A. Let ae mol / lit of PCI, be dissociated PCI, = PCI, + Cl, Initial conc : 1mol/lit 0 0 | Lquilibrium conc : (1—x) mol/lit x mol/lit x mol/lit | K, = 0.0211 moVlit [PCH }[C/] XXX ilibri K, =————+ >0.02ll= Equilibrium constant, [PCi,] (=x) => 0.0211-0.02211x =x? =x? +0.0211x—0.0211=0 if 0.02114 (0.0211)? - 4x1 (0.0211) _ ~0.0211+0.2913 _ -0.0211+ 0.2913 2x1 2 . 2 (as value of ‘x’ cannot be negative hence we neglect that value) = 0,135] The equilibrium concentrations are [PC/,] = 1 ~ x = | — 0.1351 = 0.8649 [PC/,] =x =0.1351 [Cl] =x =0.1351 =>x=-  a a - vine air “ae oh x , 6. For the reaction A+B=23C at 25°C, 03 litre vessel contains 1, 2, 4 mole ofA, 8 ang | C respectively predict the direction of reaction if a) K, for the reaction is 10 b) K, for the reaction is 15 ¢) K, for the reaction is 10.66 1 AL A+B 2C; [A] = mol / tit; [B]=4 mo i, ic] raal/ tit sy) Reaction quotient, Q2 {cl = oe = 84 nt = 10.66 ([A][B] (1/3(2/3) 27° 2 a) K.= 10 and Q. = 10.66 WK<Q So the reaction proceeds in reverse direction. b) K.=15 and Q. = 10.66 “K>Q So the reaction proceeds in forwards direction. c) K.= 10.66 and Q.=10.66 ., K,.=Q So the reaction is at equilibrium 7. A mixture of H,, N, and NH, with molar concentrations 5.0x10~ mol L "4.0 « 10° mol L* and 2.0 x 10° mol L" respectively was prepared and heated to S00K. The value of K, for the reaction : SHyy) + Nig = 2NH,,) at the temperature is 60. Predict whether ammonia tends to form or decompose at this stage of concentration, A. [NH,] = 2.0 x 10° mol.L*; (N,] = 4.0 x 10? moL.L"; [I L] = 5.0 x 10 * mol ' Reaction quotient, Q_ for the reaction Nog) +3H yg, = 2NH,,,, is, (NH, (2.0x107)? Q =a =——— * __ = 809x103. x pr [N,[H,> (4.0x10°)(6.0x10°) re ee ceo) Since Q. > K,, the reaction will go in the left direction and ammonia will decompose. 8. At 500K, kK, value for the reaction: 2S0,,. +0... = 2SO,,,, is 2.4x 10". Find the value of K, for each of following reactions at the same temperature. a) SO,.,+1/20,¢, SOx. b) SO,,, =S0,,, +1/20,, €) 380,¢+3/20, = 380 | A. 280.) + Ox) = 280, K,=2.5x 10" | ) 8) SOrg) + 1/20) == SOx) K, =(K,)!? =(2.5x10")!? =1.58x10° b) §0,9 se 80,5 + 120.5 St aan ot | I Mi Tsex0? €) 380.4) + 9/20.) == 38054) : K, = (K,)"? =(2.5x10")?? =3.95x10" ee | eamemeere BS SET A itt nn ree 9. K_ for the reaction N,O4g) = 2NOxq is 4. 63 x 10° at 25°C i a) What is the value of K, at this temperature. b) At 25°C, if the partial pressure of N,O,-,, at equilibrium is 0.2 atm, calculate equilibrium pressure of NO,,,, ) : ) 5 A Nig SE ANOy, «AMM, 2-11 K, = 4.63x10 s R = 0.0821 lit. atm mol'!. K"! T= 254273 =298K a) Kp =K RT)" = 4.63 x107°(0.0821x 298)! =113.28x107 =0.11328 atm b) Py o, = 0.2 atm 20 | : 5 | P; | K,=—%2 = 0,11328= - = Rio, = 0.11328%0.2=> Py, = V0.11328%0.2 = 0.1508 atm N,Q, 10. At 27°C, K, value for the reversible reaction PCI... = PCl,,, + Cl, is 0.65, calculate K.. A. K,=0.65; R= 0.0821 lit. atm. mol'K7; T=27 +273 = 300K PCI, = PGue, + CF An=n,—n, =2-1=1 348) 2(g8) K, =K,(RT)” = 0.65 =K,(0.0821x300)' > K, = — 965 __ _ 9.9264 mol / tit 0.0821x300 11. K, for the reaction, N»., + 3H — 2NH3¢) is 0.5 at 400K, find K,. A. K, =0.5: R = 0.0821 lit.atm. mol! . K"'; T = 400K Nag + 3a) = 2NA yg) An=n, ~My = 2-4=-2 0.5 —_—" ___ = 4.636« 10+ atm? (0.0821 x 400)? K, =K,(RT)™ =0.5(0.0821= 400)? = 12.1 mole of A and 1 mole of B are taken in a 5 litre flask, 0.5 mole of C is formed in the equilibrium of A+B —C+D. What is molar concentration of each species if the reaction is carried with 2 mole of A, | mole of B in a 5 litre flask at the same temperature. A. a) a + B = C+D Initial ; 1 1 0 0 Moles at equilibrium: (1-x) — (1-x) x x but n, = x = 0.5 (i-0.5) (1-0.5) 0.5 0.5 =05 =0.5 _ {CJD} _0.1x0.1_, 0.5 [A] =[B]=[C]=[D]=—— =0. i F x (A]=[B]=[C]=[D] rs 0.1 mol/lit K, [A][B] 0.10.1  arpa accel ws ee ee . th ™ £ 17.A vessel 9! 100K contains Co, with a pressure of 0. 5 atm. Some of ag co, isT the value of K, if total converted into CO on addition of graphite. Calculate | __ pressure 2 equilibrium is 0.8 atm. \ A. CO4g, + Coyne 2COG, Initial pres ure: 0.5 atm 0 atm Equilibriun .. pressure: (0.5—x) atm 2x atm Total press ire, P= 0.8 atm But, P=P, 5, + Poo =>0.8=0.5-x+2x x 20,3 Equilibriun pressure are: P., =0.5—x =0.5-0.3=0.2; P., =2x =2x93=0.6 2 — Peo _ (0.6) =1.8 atm ~ "Poo, 0.2 18. The K, value for the reaction H,,)+1,4) = 2HI,, at 460°C is 49. If the initial pressure o° H, and I, are 0.5 atm respectively, determine the partial pressure of each gases at equilibrium. A. Fag + ly = 2Hly Initial pres: ure (atm): 0.5 0.5 0 i Equilibriun: pressure(atm): (0.5—x) (0.5-x) 2x K, = 49 [Given] 2 2 > 2 Gee ae OP ee Sn 35-95 Pa x P, (0.5 —x)(0.5 — x) (0.5—x) 0.S-—x So Ke bd = 0.388 9 19.0.5 mol of H, and 0.5 mole of I, react in 10 litre flask at 448°C. The equiliyrican | constant k _ is 50 for Hy. +14, = 2H. ) a) Whati' the value of K, b) Calculate the no. of moles of I, at equilibrium A. a) Hyg + 2%) = 2H, An=n,-ny=2-2=0 K_=5S0 K, =K, (RT)*"= 50 (RT)°= 50 b) Hag + lg = 2H) Initial r oles: 0.5 0.5 0 Mole ai equilibrium: (0.5-x) (0.5—x) 2x 0.5-x 0.5-x 2x Cone, «: equilibri i = ‘onc, «: equilibrium (mol/lit) 10 10 10 “wy 2 ‘Ack a aan. (2). 10 i Oe Se C410,) 10) (0.5-x) (0.5-x) (0.5-x) BD» % Sa~ —= 50 =7 => 2x =7(0.5—x) = 2x =3,5-7x 39x =3.5 33 3x = =<*= 0.389. No.of moles of I, =0.5 -x=0.5 - 0.389 = 0.111 a | 2 | | ! : A. is ‘congaatvation of 0.1 “mote of Cl, at apelin, Ke for PClag et PC hy ot Ch (9) (g) 0.0414M. s PCisig) F2 PCH, + CL ds) initial cone: a 0 0 Cone. at equilibrium: — a ~ x x x But [C/,] at equilibrium = x = 0.1 Cain jum concentrations of gases are ; {PCi,} =a -x=(a-0.1) ([PCLJ=x=0.1; [Ci]=x=0.1; K,=0.0414 Equilibrium constant, K, oc aheia [PCi,} 0.1x0.1 = POD = 0.0414a-0.00414 = 0.01 a-v. => 0.04140 = 0.01414 eas xi3415 0.0414 Amount of PC/, added = 0.3415 moles. 2. X for the reaction N,.) + 3H, = 2NH,,, at 400°C is 1.64 x 10+. a) Calculate Ke b) Calculate AG° value using K, value. A, a) Nag + 3Hig = 2NHyQ) An =D, —Mp =2-4=-2 R =0.0821 lit. atm. mol’ K7; T=400+ 273 = 673K K = 1.64x 10% Ko =K(RT)An 1.64 x 104 = K (6.0821 x 673)? oK, = £.64x107 (0.0821 673)? = 0.5006 mole? lit? b) AG =-2.303 RT log K, =—2.303x8.314x 673 log(0.5006) = 21893.35 J 22. Calculate the pH of a) i0° M HCI b) 10°M H,SO, c) 10*MHNO, d)0.02MH,SO, A. a) HC/ is a strong and monobasic acid. * [H*] = Molarity x basicity = 107 x 1 = 103 pH = -log [H*] = -log,, 10° =—(-3) log,, 10 =3 b} H,SO, is a strong and dibasic acid {H*] = Molanty x Basicity = 10° x 2 =2 x 10° pH =—log,, {H‘] = - log,,(2 x 10”) = -log,, 2~ log, 10°? = -0.3010 ~ (-3) tog, 10 = 3 - 0.03010 = 2.6990 c) 1NO, is a strong and monobasic acid. Cry = ne x oe 10°x 1=10° pH =-log,, [H"}] =--log,, 1 $= — (-6) log, 10=6 ee d) H,SO, is a strong and dibasic avid. SS ie ee ~. [H*] = Molarity basicity * 0.02 «2 = 4 « 10? pH =- log,, [H} = - log!0(4 © 10%) = = logl04 - log 10.2 =~ 0.6021 (2) log, 19= 20,6021 = 1.3979 | 23. Calculate of pH for a) 0.001 M NaQH b) 0.01M Ca(OH), ! Kc) 0.0008 M Ba(OH), d) 0.004 M Naont | A. a) NaOH is a strong and monoacidic base. | . [OH] = Molarity x Acidity = 0.001 x 1 = 103 pOH = -log,, [OH] =~ log,, 10 = -(-3)log,, 10 = 3 But pH + pOH = 14 pH = 14-pOH=14-3=11 b) Ca(OH), = Ca’? + 20H 0.01 2 x 0.01 = 0.02 [OH] = 0.02 = 2 x 107 pOH = -log (2 x 10) pH = 14 - 1.699 = 12.301 = ~(0.3010 — 2) = 1.699 c) Ba(OH), = Ba‘? + 20H” 0.0008 2 x 0.0008 = 0.0016 [OH-] = 0.0016 = 1.6 x 10 pOH =-log [1.6 x 107] pH = 14 - 2.7959 = 11.2041 =3-log 1.6 = 3 — 0.2041 = 2.7959 d) NaOH = Na*+OH> [OH] = 0.004 =4 x 10° 0.004 0.004 pOH = - log (4x10°} =3-log4=3-0.602! pOH = 2.3979 pH = 14 - pOH = 14 — 2.3979 = 11.6021 .24. The pH of a solution is 3.6 calculate H,O* ion concentration. A. pH = 3.6 :. [H,0*] or [H*]=10°°" =10°° =10** =10%* x10 =Antilog0.4x10~“ =2.512x10~ moic/lit ‘ 25. The pH of a solution is 8.6 calculate the OH’ ion concentration. A. pH = 8.6 “. pOH =14-pH =14-8.6=5.4 [OH-]= 1Q7PO# = 1954 = 10°58 =10°° x10 =Antilog (0.6)x10~ = 3.981x 10° moles / litre | 26. What is [H*] for a solution in which a) pH =3 b) pli =4.75 c) pH=44 A.a) pH=3 0», [H"]=10 =10° b) pH=4.75 +. [*]=107" =10%" =10° 5 =10*"5 =Antilog(0.25)x10° =1.778x10* mole/lit c) pH=4,4 «[H*}=107" = 10-4 = 10° = 10" «10° =Antilog (0.6) 10° = 3.981 x10 * molelit