Question

Aniline (C6H5NH2, Kb=4.3 x 10^-10 at 25C) is an industrially important amine used in the making of dyes. Determine the pH of an aniline solution made by dissolving 3.90g of aniline in enough water to make 100mL of solution

Answer 1

e number Molar concentration of Molarity (M) is defined as the numb of moles of salute dissolve in one liter of solution. Therefore, Moxarity (M) = - Number of moles of solute Literiss of nolution. Again, Number of mole of solute = - Mano - Molari mana Given volume of nolution = 100 mL = 0.14 Bath, amone of Aniline, C H5NH2 = 3.909 Now Atomic mana of c = 12.0107 glmol Atomic maps of H= 1.00784 g/mol Atomic maroc of N = 14.0067 gimol Therefore, Molari mana of Aniline, Co HG NH₂ = 6 (12.0107 g/mol)+7(1-00784 glmol) +1(14.00 67 gmol) - 72: 06 42 glmok + 7.05488 g/mol + 14.0067 g/mol = 93.13 g/mol 3.909 Then, number of mole of aniline, Co HG NH₂ = 93.13 g/mol = 0.0419 mol 0.04 19 mot Therefore, molar concentration of the solution = O.IL = 0.419 mol/L = 0.419 M for aniline -10 tot Again, givenky = 4.3 X 10 at 25c so, pkb = -log ko = - Log ( 433 x 1õº) = 9.37

Aniline is a weak base. For weak bare, we can write, pOH = 1 pkb - 12 Loge = '(937) –vs Log (0.419) = 4:685- (-0.38) = 4.685 + 0.19 = 4.875 At 25°C, PH= 14- POH = 14- 41875 = 9.125 Therefore, the pH of the aniline solution is q. 125