Question

answer is 3.8g

35. What is the amount of ammonium chloride that has to be dissolved in 190 mL of 0.189 M NH3 (Kb 1.8x10 5) to make it a buffer with a pH of 8.96, in grams?

Answer 1

Amount of ammonium CHoride = ? Molarity of Nilig solution = 0.189 M to for NH3 = 118x10-5 pkb = -log ko = -log [ 1.8 x 105] = -[ log(1-8) + log 100] 3- [ 0.2552 - 5) - - (-4.7448] pk = 4.74 pH pol of the buffer solution = 8.96 of the buffer solution = 14-8.96 = 5.04 According to Hendeym - Hasselbalch equation. poh - pkt + log [ salt] [Base] poh - pko + log [Ammonium Chloride] [Amonia] 5.04 = 4.74 + Loy [ Ammonium Chloride) 0.189 5.04 -4.74 = log (Ammonium chloride] - log (0.189) 0.3 = log [ Ammonium chloride ]- (-07235) 0.3 = log [ Ammonium Chloride) + 0.72.

-0.42 = log[ Ammonium Choside] [ Ammonium Chloride) = jó De4 [Ammonium Chloride) = 0.3801 M The concentration of salt rie. Ammonium chloride = 0.3801 M Volume of solution = 190 ml = 0.194 Molarity = No. of moles of Ammonium Chloride volume of solution (in 2) Molarity Molarity = Mass of Ammonium Chloride Molar mass of Ammonium Chloride X volume (inc) 0*3801 = Mays of Ammonium chloride 53.491 g molt x 0.192 ammonium chloride = 0.38018 53.491 X 0.19 = 3.8630 g So, Mass of ammonium Chlorid grequired is 3.89