Question

5. Calculate the [H ] of the following mixtures. (2.5 marks per part) a. Dilution and Common lon problem: 25 mL of 0.10 M HCIO, and 25 mL of 0.15 M HCIO (K 2.9 x 10* b. Buffer problem: 25 mL of 0.025 M lactic acid (for which K of 0.015 M NaOH 1.37 x 10) and 25 mL c. Dilution and Weak Acid problem: 25 mL of 0.10 M ascorbic acid and 25 mL of water d. Dilution and Conjugate Acid problem: 25.0 mL of 2.50 x 10 K 1.05 x 10 and 50.0 mL of 1.25 x 10 M HCi M nicotine (for which

Answer 1

a) Data 25 mL HCIO4, 0.10 M 25 mL HCIO, 0.15 М K HCIO 2.9x 108 We consider that the volumes are additive: V, 3D V нсIо, + VнсIо %3D 50 mL %3D 0.05 L We calculate the n(moles) of H,0* in the HCIO4 solution Dissociation of HCIO (strong acid) CIO4 НаО (а) HCIO Н,ОФ P4 (aq) 4(aq) n H,0 in the HCIO [HCIO 0.10] M = [H30] пно- 0.10 mol Kx 25 2х 1 1000 pL 2.5 x 103 mol n H,0

Dissociation of HCIO (weak acid) H20 НО (а) HCIO CIO (aq) (aq) Initio 0.1 0 0 0.1 x Equilibrium X х Applying equilibrium expression: К, - [.O©HILOol [HCIO] x2 К, 0.1 x We consider 1 - X = 0.1 since the K <<103 x2 Ка 0.1 K3 (0.1)X2 2.9x108 (0.1) x2 2.9x109=x 5.39 x 105 M [H20'] x = 5.39 x 105

n Ho in the HCIO 5.39 x 10 M = [H,0'] = 5.39 x 105 M [HCIO] n H,0 5.39 x 105 mol / Jx 25 ml x 1 1000 mL n H,01.35 x10 mol n, = 2.5 x 10 mol + 1.35 x106 mol n = 2.5 x 103 mol [H] of the mixture is [H] = [H,0 [H*] V 2.5 x 103 mol [H*] = 0.05 L [H*] 0.050 M

b) Data: Lacti acid (C,H,OHCOOH): 25 mL (0.025 L), 0.025 M (K = 1.37x104 NaOH: 25 mL (0.025 L), 0.015 M [H]? Assuming that the volumes are additive, the total volume of the mixture is: + V. NaOH Vtotal lactic acid Vt 0.025 L + 0.025 L total V, 0.05 L total

Neutralization: С,Н,ОнСоONa + H,о С,Н,Онсоон + NaOH Lactic acid Sodium Lactate n NaoHM x V n NaoH0.015 mol /Kx 0.025 K n NaoH 3.75x104 mol =ncH,OHcooH neutralized n CHOHCO ON a

Calculation of the concentration of sodium lactate formed 3.75x104 mol [C Н.онсоо] - 0.05 L [C2HOHCOO 0.0075 M Calculation [H] of the buffer solution formed: С,Н,онсоон + но С,Н,ОнСоо + H,о Initio 10 0.0075 0.0075 x Equilibrium х х [с-Нонсоо" ][н'] К, a [с,Н.онсоон] 1.37x104 Approx x << 10 0.0075 x 0.0075 - x 0.0075 х? 1.37x104 0.0075 1.37x10(0.0075) = x [H*] 0.001 M 0.001 = X