Question

1. If you needed to perform a reaction in a controlled pH environment that was fairly basic, you might choose to use the ammonium and ammonia buffer system. How many grams of solid ammonium chloride would you have to add to 250 liters of 0.165M NH3 to obtain a buffered solution of pH = 8.85?

(K_b of NH₃ = 1.8 x 10^-5)

Answer 1

Given :- Concentration of NH₃ = 0.165m Volume of NH3 = 250 litres pH of buffer solution 58-85 Kb of NH₃ = 118010 1.8 X/ Solution : Buffer is a solution which resists changes in pH when an acid (or) alkali al water added to it. Ammenical buffer is made up of addition of ammonium chlorido (conjugate acid) to' ammanca Selection (weak base) NHD + NHO Ammanical buffer Conjugate weak Vacid base Hendessen - Hasselbalck equation is used to find out the concentration of Conjugate acid . POH = pk + log/{NHa* pot = pkb 1 [NH 3] pH = 8.85

pOH = 14- pH pOH = 14 - 8.85 [pOH= 5:15] Step (li) calculate pro рк, -а, К. - log (1.8x105) = -log 1:8 + 5logio -0.2553 + 5 4.7447 5.15 = 4.74 [pky = 4.74 Slap (i) pOH = pkg + laq (Cohen 5:15 = 4:74. + lag [ what] ) 1 [nHg] 0.41 = log (CNH] To get aid of the log, use [NH 4t 0.41 - 10 [NHz

We get, J: 2.5704 {NH3 This gives the ratio between the concentration of NHAT and concentration of NH3 [NH4+] = 2.5704 . [NH 3] Sub., [NHg] = 0.165M , then | [NH : 2-5 104 (1-1.5 M) {NH] = 0.4241M Ammonium chloride dissociates completely is aqueous solution to give ammenumion and chloride ion NHAC (ag) Nh (aq) + dog) I mel of whad peoduce I mad of NA ion. {NHACIJ = {NHat] = 0 4241 M Assuming that the volume of the solution

does not change after the number of moles solution is calculated Moles n= addition of NHA of NHcl in the using the formula, Where C - Concentration : 0.42414 V = volume = 250 L Bermula n = 0:424.1 mol 14 x ( 250 %) n = 106. 025 moles NHAI Step (iv) Calculate how many grams a NHA I required uring the po Moles n Mass (m) Molor mass (M) Moba mass of NHAU = 53.49 g/mol Mass = nxM = 106.025 stel * (53.49 gloke) - 5671.29739

5671g of solid ammonium chloride add to 250 liters of 0.165M NH₃ to obtain a buffered solution of pH = 8.85