Question

From the book of Chemistry, The Central Science 12th edition. Chapter 15 and 16 answer the following question. B) Calculate the pH of an aqueous solution of ammonia, NH3(aq), 5.00 x 10-3 M. Follow the steps below to solve the exercise considering Ammonia is a weak base. The basic constant for ammonia is Kb = 1.8 x 10^-5 1) Write the dissociation or ionization reaction of ammonia in water 2) Indicate in a table the initial molar concentration, the change in concentrations and the equilibrium molar concentration below the chemical species. 3) Write the mathematical equation that expresses the equilibrium constant and substitute equilibrium concentrations. 4) Calculate [OH-] using an approximation where x <<< 5.00x10-M 5) Verify the approach using an acceptance criteria of 5.0% 6) Calculate the POH 7) Calculate the pH

Answer 1

CD Ionization reaction of Ammonia . NH₃ laq) + H₂011) NH+ (aq) + On (aq) Draw the I-c-E table NH3 laq) + H₂0 (1) NHut (aq) + On lag) 58103 m 0 o Initial change Equilibrium -a 42 ta (5x10 3n) e n (3) The expression for Kb can be written K, - [N+] [or [NH₃ Pulting the rames at equilibrium, we get. 1.8 x 10-5 - [x] [n 49 [5x10-3_n 53

1.8 x 105 [2] [m [5x103 n 1.8x10-5 x2 (5x103_n) (4) We to assume that Said n<<< 5810-3 equation can be the above approximated as 1• 8 x 10-5 = 22 ÿ n → (since very small) is 5x10-3 108x5 x 10-5-3 2 >> 2² 8 22 a²= 2 1.8 x5 x 10 9 x 10-8 + 9x108 € 3x10-4 a n Neglect the negative value of n beaue concentration can't be negat gatve. 50 53 Com ] n = 3x10-4 M

(6) If we that to do not take the assumption nck Sя ID-3 м LOе. have sowe the quadratic equation 1-8x10-5 x2 (5x103_n) 2 9x10-8 O 11 n²+ 9x108 solving we get, On 2 2.9x10-4 ... 1. error is - X100 Enact - Appron- Appron. | (3-2-9) Xeory 3x184 O.1 x 100 -- X 100 3 501: 3.33% which is less than acceptance criteria. So the assumption is 51 53 طا Teaserable.

(6) ярн — Іоа Сон-1 Соя L3xis" 1] Яън 3:32 РОН E (1) РИ + РОИ 14 pH = 14- 3:32 Ph= 10. Ч8