Question

1. What is the pH of a buffer consisting of 0.30 M CH3COOH & 0.20 M NACH:COO? Ka= 1.8 x 10-5 2. What is the pH of the buffer with 0.10 M NH3 and 0.20 M NH4NO3? Kl = 1.8 x 10-5 3. What is the pH of a buffer formed from combining 10 mL of 0.250 M HCl with 90 mL of 0.150 M NH3? (K = 1.8 x 10-) 4. Calculate the pH of the solution that results from mixing 55.0 mL of 0.15 M HCOOH with 80.0 mL of 0.12 M NaHCOO (K. - 1.8 x 10)

Answer 1

D one of weak aud CH3COOH = 0.30M Cone of conjugate bave CH3 CO NA = 0.20 M ka= 1.6x10-5 As we know pka - dog (ka) = – log (1.8x103) -- [log 18 + log 105 – (0.26-5) - 4.74 From Henderson- Hasselbalch egr pH = pka + log [conjugare bose] weak and I By substituting all valles pH = 4.74 + log ( 0.20 pH = 4.74 -0.16 IPH= 4.56 0.30 (2 cone of weak base= 0.10M cone of conjugate aud = 0.POM Kb = 1.8x10-5 Pkh= -log kb = -log (1.6x105) pk = 4.74 : foom Menderson. Harrel balch egalen POH= pkb + log (conjugate aud] weak Base] - 4.74 + log ( 0.20 ) = 5.04 0.10

1. PH+ POH= 14 14- POH = 14-5.048.96 º Ha + NH3 (stoong aud) (weak bose) + NH4 ( te Hu +P HUP NH3 react in a 1:1 mole ratio. This means that we need equal amount of moles of each reactant in order to get a complete neutralization. conc. of HU = 0.250M in volume of Hell uned = tome No. of moles of Hu= cone-avoll in its) = 0.25ox 10 = 0.00250 0 1000 (000 cone of NH3 = 0.15OMI volume of NH3 used = 90ml & no. of moles of NH₃ = Concx vollin Hr) -0.150 X 90 = 0.0135 since no. of moles of NH3 is more than HC, which indicates that Hu in acting as limiting reagent 4 it will be completely consumed in the grin - 0.009 50 moles of HU will react with 0.00250 moles of NH3 to produce 0.00250 moles of NHY. : moles of NHy = 0.00250 . After the completion of thing NHU = 0 (where n number of moles) M NH3 = 0:0135 -0.002 80 = 0.011 ĐH L = 200 250

0.00250 100x40-3 total volume of 10th= 10 +90= 100ml io cone of [Nitty ] - moves volumel in ltr = 0.0250M K6 = 1.6x10-5 pk = 4.74 i POH= pkb + dog [conjugate aud] [weak base] conc. of weak base NH3 = 0.011 11M 100X10-3 POH=4.74t dog (010250 0.11 POH= 4.74 - 0.64 = 4.1 pH = 14-pOH = 14-4-4-9.9

20 ka = 1:8x104, : pka = 4:74 cone of HOOH = 0.15M volume of Hook used=550mL un no. of moles of HCOOH = Conc. a vollin (tr) - 0.15X55 – 0.0075 1000 tone of base HooNa = 0.12 M. volume used= 80.ml neay no. of moles of twoNa = 0.12160 = 0.0096. 1000 PH - ple + (1ң сор на hool и : 3 - 4.74+ dog 0.0006) ph= 4.74 - 0.46 TpH = 4.28