Question

4. When 0.100 mol ammonia is dissolved in water to make a 500 mL solution, the following equilibrium is established NH3(aq) + H2O(0) - NH. (aq) + OH(aq) K = 1.8 x 10-5 a. Find the concentration of OH' and the pH of the solution at equilibrium. Remember: pOH = -log(OH) and pH + pOH = 14 b. Consider the following reaction: 2H2O(1) 2 H,0*(aq) + OH(aq) K = 1.0 x 10-14 Write the equilibrium expression and find the value for the following reaction: NH, (aq) + 2H2O(1) 2 H,0*(aq) + NH3(aq)

Answer 1

6 solution - given, mol st volume of amonia - ol mol solution - soonl. mol of HH3 x10oo. volm of sol (mi) amonia - - concentration of 0.1 soox 190o - 02 M. given Reaction's NH₃ (49) + H₂ oll) Initial conc. 0.2 Nhy (aq) + tot (49) K = 1.8x105 conc. at 0.2-x. equillibrium: (NH4] [ott K = xxx sy 1.8410 = -0 2-0 - - => i x = 0.36X105 – 1.8x10 . x + 1.8x10 – 0:36x105 = o. bee = 1:888 X 10-3 rusing Scientitic calculator] concentration of st - 1.888*10 M. pot – – 10g [opt = -19 (1.888X109) - 272 PH+ oH = 14 & PH - 14-pot - 14-2.72 now, PH = 11:28

pH of me solution - 11.28. giver, 2H20 (1) ² Hz or at OH (as) Ke=1x1014. K = [H₂ ot] [ot] = 1014.04 [1904] - Ton] - NHy (21) + 2H20 - Hot (as) + NH3 (as) K. - [H₃O+] [NH₂ [Nhy 1 * [ty ] (trom-] [NH 10-14 7 x [H Con I x 1014 from - Vac K= [NHyt] [ot] I NH₃ 1.88 los 8 xlors X10-14 5.55*10-10 the Value of equillibrium constant (ke) for above reaction is 555 x 10-10