Question

Question 1 HA is a weak acid. Its ionization constant, Ky, is 5.1 x 10-13. Calculate the pH of an aqueous solution with an initial NaA concentration of 0.053 M. Question 2 We place 0.134 mol of a weak acid, HA, in enough water to produce 1.00 L of solution. The final pH of the solution is 1.18. Calculate the ionization contant, K. of HA. Question 3 We place 0.577 mol of a weak acid, HA, and 13.6 g of NaOH in enough water to produce 1.00 L of solution. The final pH of this solution is 4.50. Calculate the ionization constant, K. of HA.

Answer 1

Soi NGA » Not ta 1.53 0:53 Here is Salt, so Hydrolysis is A + H₂O ОН HAT 0.53 D53-1 there. 0.53-u {osz>>>] 5-181013 22² ㄷ 0-19 X).-) 0,53 = na 10.57X0.19 Xiot こ 0-103 x 10-1 님 1-01 X10-4 =fon - 17.03 x102 = loglout log - Isgl.ol pu pouz Otx107 pon Z o-996 14-0.956-

+ Sol Kane 2 ca 1-a HA и 0.134 zu Ka 2 a 0.134 - ca kaz 22 0.1344-2a 2 (6.7x102,² 1-18 Ka Z {om24 ->>> 2 pha - Hylt) 2118 Lt) = lo 6: 134 Ela-1-18 747) = 6-74102=ne 3.2xion