Question

NH4+ salt is confirmed

8/24 points Previous Answers My Note Calculate the change in pH to 0.01 pH units caused by adding 10. mL of 2.82-M NaOH is added to 390. mL of each of the following solutions. a) water pH before mixing = 7.00 pH after mixing= 12.85 pH change = 5.85 b) 0.127 M NH41+ > pH before mixing = 5.07 pH after mixing= 9.50 X pH change = X c) 0.127 M NH3 pH before mixing = pH after mixing= pH change = d) a buffer solution that is 0.127 M in each NH41+ and NH3 pH before mixing = pH after mixing= pH change =

Answer 1

b) 0.127 M NHAU Kb of NH3 = 1.8 x 10-5 K 1x10-14 K, of NH K 1.8x10-5 = 5.56x10-10 The dissociation of NH4 is given by К. | NH, NH4 + H+ Initial 0.127 Change -X Equilibrium 0.127 - X K. X.X =5.56x10-10 127-X .x is small, so 0.127- x=0.127 => x = 0.127x5.56x10-10 = 7.06x10-11 x=8.40x10-6 ::[H*]=8.40 x10M pH = -log[H*]=- log (8.40x10“)=5.08 pH of the solution = 5.08

Addition of NaOH: Number of moles of NH4+ =M*V = 0.127 M * 390 mL = 49.53 mmol Number of moles of NaOH = M*V = 2.82 M*10 mL = 28.2 mmol Net moles of NHat remaining = 49.53 – 28.2 = 21.33 mmol Total volume of the solution = 390 + 10 = 400 mL - Ka NH+ + OH = NH3 + H2O 49.53 0 Initial Add Add 0 28.2 28.2 Change - 28.2 - 28.2 +28.2 Equilibrium 21.33 0 28.2 pH of the solution is determined by Henderson's equation [base] pH = pka +10% acid] => pH = – log(5:56x103") +log 28,35 = 9.255 +0.121258= 9.38 pH change = 9.38 – 5.08 = 4.30

NH,+: pH before mixing = 5.08 pH after mixing = 9.38 pH change = 9.38 – 5.08 = 4.30 c) 0.127 M NH3 Ko of NH3 = 1.8 x 10-5 The dissociation of NH3 is given by 1 Ko NH3 + H20 = | NHA + Initial 0.127 Change -X X.X Equilibrium 0.127 - X K; = 1 =1.8x10- 0.127- x? -1.8*10*x+2.29x10“ = 0 by solving we get, x=0.001503 :: [OH]=0.001503M POH = -log[OH-]=-log(0.001503)= 2.82 pH = 14 - POH =14 – 2.82 =11.18

Addition of NaOH: Number of moles of NH3=M*V = 0.127 M * 390 mL = 49.53 mmol Number of moles of NaOH = M*V = 2.82 M*10 mL = 28.2 mmol Since both NH3 and NaOH are bases. They will not react with each other. NaOH is the dominant producer of OH ions. [NaOH] = moles/total volume = 28.2 mmol/ 400 mL = 0.0705 M (NH3) = 49.53 mmol / 400 mL = 0.1238 M NH3 + H,0 = NHa* + OH Initial 0.1238 0.0705 Change -X + X Equilibrium 0.1238 - X 0.0705+ x r(0.0705+x).x, 0.1238 - x -=1.8x10- 0.0705x + x2 = 2.23x10-6 -1.8x10-* x x? +7.052x10-2 -2.23x10-6 = 0 by solving, we get r=3.16x10-5 :: [OH"]=0.0705 +x=0.0705 +3.16x10-* =0.070532M pOH = -log[OH-]=-log(0.070532)=1.15 pH = 14 - pOH =14–1.15 = 12.85

NHz: pH before mixing = 11.18 pH after mixing = 12.85 pH change = 12.85 - 11.18 = 1.67 (D) Solution containing both NHz and NHA pH of the solution is determined by Henderson's equation pH = pKa + log [base] [acid] 0.127 pH = -log log =9.255 +0 = 9.26 0.127

Addition of NaOH: Number of moles of NH+ =M*V = 0.127 M * 390 mL = 49.53 mmol Number of moles of NH4+ =M*V = 0.127 M * 390 mL = 49.53 mmol Number of moles of NaOH = M*V = 2.82 M*10 mL = 28.2 mmol + - NH3 + H2O NH+ 49.53 Initial OH 0 28.2 49.53 Add Add 28.2 Change - 28.2 - 28.2 +28.2 Equilibrium 21.33 77.73 pH of the solution is determined by Henderson's equation pH = pKa +log [base] => pH = – log(5:56x103") +log 77.33 – 19.255 +0.561598=9.82 =pH =-lo 77.73 21.33 =9.255 +0.561598 = 9.82 pH change = 9.82 – 9.26 = 0.56

(d) For NH, and NHAT : pH before mixing = 9.26 pH after mixing = 9.82 pH change = 9.82 - 9.26 = 0.56