Question

A solution (75.0 mL) of 0.250 M NH3 is being titrated with 0.500 M HCl. Kb = 1.8x10–5 for NH3. What species are present, and what are their concentrations:

(a) Before any HCl has been added?

(b) When 17.5 mL HCl have been added?

(c) After 37.5 mL HCl solution has been added?

(d) After 45.0 mL HCl solution has been added?

Can you please show all work?

Answer 1

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Given mmol of NH3 Somolarity xVolumelinmiy = 0.250 x 75 ml = 18.75 mmol Ko = 1.80x105 pky s log ky = -log 1-8x105 -4.74 as Before any HCI has been added - | NH3 + H2O NHA + OH I (M) C Com -ca +ca +64 E(M) C-cx ca ca Inmutoh? - qaxca = ca ² Sci- 1- as tb is very very less (1-) 1 k, = 042 67 as 1-9x10-5 - 8.49x105 0.250 os [on] - Ca = 0.25 X8.48x103 M = 2.12 x 10 ºm - pora -log [on] = -log 2012x10-3 = 2.67 Let = 1x -Poh = (x - 2.6 7 8 Lị:33 ---PH=11:33 bo mmol of Hol added > 0.50X 17.5 = 8.75mmal NH2 + HCI - NHACI Total volume I (mmoll 18.75 8.75 = 75 +7.5 -8.75 -3.75 +9.75 = 92.5ml | E(mmol) 10 8.75 forms a buffer solution whose pH calculated by pok= pkb tlog salt 4-74 + log/8.75/92.5 10/32.5 POH = 4.68 pH = 14 - POH = 14-4.68 = 19:32 Scanned with CamScanner

C) mmol of HCI added = 0-50*37.5 318-75 Nl₂ + Hat Nhace Himmell Total 19.75 18.15 Volume - 18.75 -18.75 +28.75 35 +37-5 3.5 mmol =112-5 mi It forms a salt which is formed weak base strong acid & ph can be calculated by pH = 1 (pkw -pki - bg c) [Nitycl] - 18.75 nmol = 0.167 112.5 ml pH = 2 ( 14-4-34 - log 0.167) PH= 5.02 d) mmol of Hal Qolded = 0-50 * 45 = 22.5 mmol NHz + HCL NHUSC I (mno) 18.75 22.5 - Total volume = C -18.75 -18.75 +18.75 75+45 = 120m2 Elmmol) — 3,75 18-75 Now pre will be defined by Hel only [hai] - 3.75 mmol - 3.125x10 2M .pH = -log [nt] = -log [ 3-125x102) (pH=1.51 Tesl Scanned with CamScanner