Question

22. + -10.1 points 0/4 Submissions Used Determine the pH of each of the following solutions. (a) 0.360 M phenol (weak acid with Ka = 1.3e-10). (b) 0.458 M hydrogen peroxide (weak acid with Ka = 2.4e-12). (c) 0.238 M pyridine (weak base with Kb = 1.7e-09).

Answer 1

(a) Given the concentration of C,H,OH is 0.360 M Ionization of C,H,OH is as follows: CH,OH(aq) = CH 0 (aq) + H+ (aq) I(M): 0.360 C(M): -x E(M): 0.360- Acidionization constant K, of CH OH=1.3x10-10 [CH,0-[H*] (C,H,OH] K. = ? 1.3x10-10 0.360- Assume 0.360 - X$0.360 since C,H,OH is a weak acid 1.3x10-10 1.3x10-10 = 0.360 to x = 0.468x10-10 x = 0.684x10-4 [H* )= x= 0.684x10-4 M ------------- -------- ------ --- -------- -- Calculate the pH as follows: pH= – log[H*] = -log(0.684x10-) = 5.16 Therefore, the pH is 5.16

(6) Given the concentration of H,O, is 0.458 M Ionization of H, O, is as follows: H,O, (aq) = HO, (aq) + H+ (aq) I(M): 0.458 C(M): -x E(M): 0.458- of H,O, = 2.4x10-12 Acidionization constant [H0,-[H] (H,] 2.4x10-22 0.458 - x Assume 0.458 - x$0.458 since H,, is a weak acid 2.4x10-4 = = 0.458 x2 = 1.1x10-13 x = 1.048x10 [H* )= x= 1.048x10 M ------------- -------- ------ --- -------- -- Calculate the pH as follows: pH= – log[H*] = -log(1.048x10") = 5.98 Therefore, the pH is 5.98

(C) Given the concentration of pyridine (CsH,N)is 0.238 M Baseionization constant K of CH N=1.7x10-9 (CsH3N] = 0.238 M Ionization of CH Nis as follows: CH,N(aq) + H2O(aq) = CH.N* (aq) + OH(aq) I(M): 0.238 C(M): - E(M): 0.238- x K, = 1.7x104 [C,HN][OH-]. = 1.7x10-9 [C,HN] x -1.7x10- 0.238-x= 1.7x10-4 Since it is a weak base, 0.238- x = 0.238 x =1.7x10 0.239 =1.7x10-4 x= V0.4046x102 x=2.01x10- [OH-]= 2.01x10- M POH=-log[OH-] =-log(2.01x10-) = 4.70 pH = 14-POH = 14-4.7 = 9.3 Therefore, the pH is 9.3