solutions are on the table, the ph values are next, then the question. thank you

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solutions are on the table, the ph values are next, then the question. thank you so much!

Table 16A.1 Composition of Solutions Solution Composition 0.10 MHC H3O2 2 5 mL 0.10 MHC Hoo, t 5 ml. Hoo 1 mL 0.10 MHC2Hvo, t 99 mL H o 5 mL 0.10 MHC1Hyo, 5 ml., 0.10 MHCI 5 0.10 M HPO 6 0.10 MNH 0.10 M NH NO. 8 50 mL 0.10 MNH,+ 50 ml., 0.10 M NH.NO, 10 mL Solution 8 6 ml. HiO 10 10 mL Solution 8 t 5 mL H2O 1 mL 0.10 MHCl 11 10 mL Solution 8 6 ml 0.10 M HCI 12 10 mL Solution 8 5 mL HO 1 mL 0.10 MNaoH 13 10 mL 0.10 MHC Hyo, 5 mL 0.10 M NaoH 14 10 mL 0.10 MNHsNO, 5 0.10 M NaoH mL,

Explanation / Answer

Solution.1 For 0.10 M CH3COOH solution the calculation for degree of ionisation can be done as

CH3COOH + H2O = CH3COO- + H3O+ (at equilibrium)

Ka= [H3O+] [CH3COO-]/ [CH3COOH] = 1.8 x 10-5 ...........EQ.1

putting values in EQ.1

1.8 x 10-5 = [x] [x] / (0.1-x) = x2/ (0.1-x)............EQ.2

Assuming that x <<<0.10

(0.1-x)= 0.1

therefore 1.8 x 10-5 = x2/ 0.1

[H3O+] = x = 0.00134

pH= - log10(H3O+) = 2.87

Degree of Ionisation = [H3O+] / [CH3COOH] = 0.00134 / 0.1 = 0.0134

Similar calculation can be done for other solutions also.

Acetic acid is a weak acid and it dissociates into the solution poorly. Therefore it provides less conc. of hydronium ion in the solution. Also it shows hydrogen bonding with water and therefore low value of degree of ionisation is shown. The difference between the calculated and observed pH value is due to random hydrogen bonding between acetic acid and water molecules.

Entity Initial Conc. Change in Conc. Equilibrium Conc. [H3O+] 0 +x x [CH3COO-] 0 +x x [CH3COOH] 0.1 -x (0.1-x)

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