Question

Consider a buffer prepared by mixing 75ml of 0.03640 M ammonia (NH3) solution with 28ml of 0.02454 M ammonium chloride solution (pKb of ammonia is 4.75)

1. What will the pH be?

2. If you add 1 mL of 0.1780 M HCl, what will the new pH be?

Answer 1

Total volume of the solution = (75+28) ml = 103 ml 75 ml 0,0364 M NH3 solution is mixed with 28 me 0.02454M Let us suppose NHAC. Now, concentration of NHy in this solution=x (M). We know V2 = Final volume =103 ul vs vs. Sv=Initial volume of NHg = 75 ml / So, 758 0.0364 = 10382 a 5.-guitial cone. of) My = 0.0364 M or x = 75*0,0364 = 0.02 65 (M) 103 orix=0,0265(14) concentration of weak base NH in solution = 0.0265 (M) So Let as suppose concentration of Hall (NHCl) in Solution is y (M) V = Total Volume V=Initial Volume of So, applying is, = V2 I sall - 28 ial Eguitial 2860.0245 concentration = 103xy of Sell_000245 Or, y = 28*0.0245 is - = 0.0066 (M) Or, y = 0.0066(M) 19 So, concentration of salt in solution is 0.0066(M) equation Now, applying, Henderson hasselbalch We get- poh = pkb+ log [sall] 1 Base

or pot = 4.75 + log/0.0066 0.0265 POH = 4.75 + (-0.6037) 08 or poH = 4.146 We know that, pH + pot = 14 So, pH = 14- pOH = (14-4. 146] =9.853 pH = 91853 pH of buffer solution is 9.853 The 12 If ime of 0.1780 M Hel is added Iue 0.178 m Hell = 1x 0.178 mol of Hel = 0.000178 mal of Hel So, 0.000178 mal of Hel will be reads with 0.000178 mol of Nty to form 0.000178 mal of NHAU. NH₃ + HCl = NHA cl 0.000178 0.000178 mol их.

Here amount of NHA ce increased by 0.000178 mal and amount of NH₃ decreased by 0.000178 mol 103 ml 0.02.65 M NH = 103 x 0.0265 mol of 1000 = 0.0027295 mol of NHg 103 ml 0.0066 M Migel = /103x0,0066 ) mal of 1800 / Nice = 0.0006798 Mol of Niger Now, amount of Nice after addition of Hee is 10.0006798 +0.000178 )imal = 0.0008578 и в Х. and amount of Nih = (0.0027295-0.00 178) mal! - 0.0025515 mal of NHA New, Volume of solution = (103+1) =104 ml. Sc, Concentration of base (NH) (0.0025515 X1600) So [NH₂) = 0.024533 (1) 104

Concentration of fall of Nigel = (0.0008578x1000) So, (sall] = 0.008248 (M) 104 Sac So, on poH = PKB+ log [ sall) [base] poH = 4.75 + log/0.008248 (0.024533 / pot = 4.75 + (-0.473402) POH = 4.27659 or or. We know that, pH + pOH = 14 or pH = 14-POH = 14-4. 27659 on pH = 9.7234 The New PH will be 9.7234.