Question

16. Each of three beakers contains 25.0 mL of 0.100 M solution at 25°C. Beaker 1 Beaker 2 Beaker 3 0.10 M NaOH 0.10 M NaHCO3 0.10 M H2CO3 a. Determine the pH of the solution in beaker 1. Justify your answer. b. In beaker 2, the reaction HCO, (aq) + H2O(1) = H,CO, (aq) + OH(aq) occurs. The value of K, for HCO, (aq) is 2.3x108 at 25°C. i. Write the K, expression for the reaction of HCO, (aq) with H2O(l). ii. Calculate the [OH-] in the solution in beaker 2. C. In beaker 3, the reaction H,CO2(aq) + H2O(l) = HCO, (aq) + H20*(aq) occurs. i. Calculate the value of K, for H,CO,(aq) at 25°C. ii. The contents of beaker 2 are poured into beaker 3 and the resulting solution is stirred. Assume that volumes are additive. Calculate the pH of the resulting solution. d. The contents of beaker 1 are poured into the solution made in part (c)(ii). The resulting solution is stirred. Assume that volumes are additive. i. is the resulting solution (after any reactions occur)an effective buffer? Justify your answer. ii. Calculate the final [HCO,] in the resulting solution at 25°C.

Answer 1

@ In beaker i:- 25 ml 0.1(M) NaOH = 0.1X25 - 2,5x10-3(M). 1000 concentration of NaOH = [Nath] = 2 so, concen ncentration of oH ion = I OH-] = 2.5x10 m 2OH = [NagH] = 2.5x15 3 (m) POH = -107 [on-] - 107(2.5x10") we know that, PH & Poh = 14 (at 25°c) pH = 14 - PH PH = 14- 2.to > PH = 11.4. Henee, 'PH of the solution in beakennis 11.4. Oil* 25 25 me 0.1(M) Natco = - looo = 2.5x15 (M) concentration of Natco z = [NaHCO3] = 2.5X163 (M) so, concentration of [Heo ;] = 2.58103 (m) w the reaction is tog (ap) + H₂O(l) H₂CO3(aq) Expression of kn [H2CO3] [ot ] hlay) [Hon] [126] [H₂ COB] [OH-] [Hcoz L H₂O)

5 f + oH (4) Initial (aq) + Hyoll Hycoplag) at of olaq Concentration (m) 2.5x16-3 At apun. (1-x).215x103 215x10 3. & 2.5810 3 d. w Where à is the des egree of dissociation k. It cop] Tot] ko (2.5 8103 &) (2.5x1532) (1-2)(215 X10-3.) As ko =2.3x10-8 (very small), so «<<1, hence na vs. 2.3x108 - (2.5810-3)*xan 2.5A163) = 2:3X108 2.5810-3 => x = 9.2x 10-6 = x – 3.03 x10-3 so, concentration of [OH-] = 2.5x103 xa = 2.5X10 3 x 3.03X103 = 7:58x10-6 (m.

@ In beaker, 3. Hylo, (a) + H2O (2) Hojla) + H₂0 (ar) Expressions - Hon] [H3O+] [Halog] 25 me. o.1 (M) H₂CO = 2500l = 2.5x18 3 (m). So, [H₂ 103 = 2.5x10-3m [Home] =2.5x 10 m and [H₂64] = 2.5x103 m (Assuming complete dissociation ) ko (2. 5x103) (2.5 103) (2.5x403) - Ka = 2.5x153 so the value of ka 25°C is 2.5x16 3. for H₂CO3 (ar) at (1) when the contents of weaker are poured into beaker 6, neither H₂ 203 (0) and Heo lag) further dissociates due to presence of common in. Total volume of the solution become . = (25+25) = 50ml, Now, concentration of [on-] = 7.85 x100 (m) coming from baxer-2) - 3.92x10-6 (M). and emeentuation of TH26] - 2.5 X10-3 ima 2 - 1:25 x16 (coming from beaker-) = 1.25 xong (m) rom be

s, in the resulting solution, there is excess. concentration of [H₂o+] = (1.258103-3192x106) - 1:24 110-3{M). resulting solution -log[43ot] S-10 (124x10-3) = 2.90 then contents of beaker 0 are poured into the solution mache in part 1 Nach will give 2.5x 10-3 (m) of of Sam the fastbirya, foneentities of th:] l The resulting solution will be an buffer of Hicon and Nation (weaxacid) it's sult) effective concentration of For beaker-2] [Hag] =(1-2) 2.5x10-3 = [(1-3.03x153) x2.5 x 16 37 3 2.4 X105.(m). concentration of THCO = 2.5x 10-3 (m) Beaker-3

Ker- & Beaker- > [H₂ 103 [Halo27 (2.4x10² + 2,5x103) 2 = 2.45 X10 3 (m). hen NaOH added to it which In resulting solution & which give [ot] in. tion & Tond 2.58 10-3x25 75. [OH-] = - 8.33x154 (m) This [OH-] will reacts with H₂CO3 lag to give Hcon, According to the equation - Haloz + H₂O Hilo + OH- (8.33x164) In resulting solution & Beaver- & Beaker-③ a Beaker - - 8.33x10.4 (from wash [H con- 2.45X16S + 8.33x154 1. 3 Bearer +3 = 8.16410-4+ 8.33 X164 Total Voim 75m [From Nast = 1064x.10.3 (M). final concentration of TH con ] at 25°C = 1.64X103 (M).
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