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Compare pH of solutions 2&4. What effect did adding NH4+ have on the equilibrium/pH?

Compare pH of solutions 2&4. What effect did adding NH4+ have on the equilibrium/pH?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Can you help me calculate the concentrations for 4&5 and explain to me what effect did NH4+ have on the equilibrium/ph. Also, can you explain to me why the pH change a smaller for solutions 4&5 then it is in solutions 1&2.

 

SOL ER Equilibria with Weak Acids & Weak Bases - Results Exp Composition 1 A few mL of 0.10M NH3 (unchanged) 2 2.5mL 0.1M NH3 & 7.5mL H₂O 3 2.5mL 0.1M NH4CI & 7.5mL H₂O 4 2.5mL 0.1M NH3 & 2.5mL 0.1M NH4CI & 5mL H₂O Take 1mL Solution #4 and add 9mL of water GEODO Iziprasidons 5 10 ml Solution 1. Use the concentration of NH3 and the measured pH above to determine the Kb of NH3. (show the ICE table and calculations) -14 PH= PKa + log ([A]/[B] [AH□ CH] 9.26- Pla+log (NHG] (NH4) 9.24 = Pleat log (a'] (0.0 Pla = 9,26 pkg==logka Ka=-antilog Pla ках кв = 10 кв = 10-14 Ka 10-14 1.81x109 Total Vol n/a 10 ml 10 ml 10 ml 1.81 M₂ = (2.5x0.1)/10 M2 = 0.025M K6=5,55 x10-24 Use the equation (Ka*Kb = 1*10-14) to calculate the Ka for NH4* ? -Antilog (9.26) = -1.81×1091 X 10 -23 1 pH 11.13 10.82 6.01 9.26 9.26 Solution 2 Use the equation M₁V₁= M₂V₂ to calculate new conc. of NH3. M₁ V₁ = M₂V2 Conc. of NH3 pH measured: 10.82 2.5 X0,1 = M2 X10mL m² = 2.5x0.¹ = 0,023M Compare solutions 1 & 2. What effect did diluting the base have on the pH? The addition of H₂0, slightly decreased the pH. Use the new concentration and pH to measure the Kb of NH3 again. Did it change? Kb will not change til the temp changes Kb=cok² x²=1,8225x105 0.025 α = 0,027 Solution 3 Conc. of NH4+ pH of solution 3: 6.01 Use the diluted concentration of NH4+ and measured pH to determine the Ka of NH4+ M₁ V₁ = m₂ V₂ 61 = 1/2 p² pk 2.5ml (0.1M) = M₂ (10m²) [OH-]=CQ=0,025x0,021 -log[04²] = -log[0,025 x = 0,027) POH= 3,17 PH=14-3₁17=1085 17 PH=pKa + log ((A] [B])/([AH] [BH]) ( KaxKb = 1x10""" 6.01 = pka +10g ( [0.137 (0.17) Kb = 1x10" / Ka pka = 6,01 X10" Ka=-1.02 x 109 Kb = 1x10"¹"/1.02X10² Kb = (1/1.02)×10 ²3 -23 Kb= 0.9804 X16-23 K5= 9.804 X10-24 9 How does th good) This Calculate th Sol 2 4 5 Compare equilibriu Com
Transcribed Image Text:SOL ER Equilibria with Weak Acids & Weak Bases - Results Exp Composition 1 A few mL of 0.10M NH3 (unchanged) 2 2.5mL 0.1M NH3 & 7.5mL H₂O 3 2.5mL 0.1M NH4CI & 7.5mL H₂O 4 2.5mL 0.1M NH3 & 2.5mL 0.1M NH4CI & 5mL H₂O Take 1mL Solution #4 and add 9mL of water GEODO Iziprasidons 5 10 ml Solution 1. Use the concentration of NH3 and the measured pH above to determine the Kb of NH3. (show the ICE table and calculations) -14 PH= PKa + log ([A]/[B] [AH□ CH] 9.26- Pla+log (NHG] (NH4) 9.24 = Pleat log (a'] (0.0 Pla = 9,26 pkg==logka Ka=-antilog Pla ках кв = 10 кв = 10-14 Ka 10-14 1.81x109 Total Vol n/a 10 ml 10 ml 10 ml 1.81 M₂ = (2.5x0.1)/10 M2 = 0.025M K6=5,55 x10-24 Use the equation (Ka*Kb = 1*10-14) to calculate the Ka for NH4* ? -Antilog (9.26) = -1.81×1091 X 10 -23 1 pH 11.13 10.82 6.01 9.26 9.26 Solution 2 Use the equation M₁V₁= M₂V₂ to calculate new conc. of NH3. M₁ V₁ = M₂V2 Conc. of NH3 pH measured: 10.82 2.5 X0,1 = M2 X10mL m² = 2.5x0.¹ = 0,023M Compare solutions 1 & 2. What effect did diluting the base have on the pH? The addition of H₂0, slightly decreased the pH. Use the new concentration and pH to measure the Kb of NH3 again. Did it change? Kb will not change til the temp changes Kb=cok² x²=1,8225x105 0.025 α = 0,027 Solution 3 Conc. of NH4+ pH of solution 3: 6.01 Use the diluted concentration of NH4+ and measured pH to determine the Ka of NH4+ M₁ V₁ = m₂ V₂ 61 = 1/2 p² pk 2.5ml (0.1M) = M₂ (10m²) [OH-]=CQ=0,025x0,021 -log[04²] = -log[0,025 x = 0,027) POH= 3,17 PH=14-3₁17=1085 17 PH=pKa + log ((A] [B])/([AH] [BH]) ( KaxKb = 1x10""" 6.01 = pka +10g ( [0.137 (0.17) Kb = 1x10" / Ka pka = 6,01 X10" Ka=-1.02 x 109 Kb = 1x10"¹"/1.02X10² Kb = (1/1.02)×10 ²3 -23 Kb= 0.9804 X16-23 K5= 9.804 X10-24 9 How does th good) This Calculate th Sol 2 4 5 Compare equilibriu Com
70° 27 b/c we are antiloy to How does this Ka value compare to the value on pg. 1? (within a power of 10 is Calculate ting value what the good) This Calculate the concentrations for solution 4 and solution 5 value is. Sol Conc. NH3 (M) Conc.NH4+ (M) 0.025M 2 0.025M pH 10.82 9.26 9.26 4 5 Compare pH of solutions 2&4. What effect did adding NH4* have on the equilibrium/pH? Compare the pH from dilution in solutions 1&2 to the pH change in solutions 4 & 5. Why is the pH change smaller for solutions 4&5 than in solutions 1&2? Think above the ratio of base and conjugate acid. You can also show this mathematically by using the Kb from part 1 with NH3/NH4* conjugates to determine what the OH- concentration should be. 2
Transcribed Image Text:70° 27 b/c we are antiloy to How does this Ka value compare to the value on pg. 1? (within a power of 10 is Calculate ting value what the good) This Calculate the concentrations for solution 4 and solution 5 value is. Sol Conc. NH3 (M) Conc.NH4+ (M) 0.025M 2 0.025M pH 10.82 9.26 9.26 4 5 Compare pH of solutions 2&4. What effect did adding NH4* have on the equilibrium/pH? Compare the pH from dilution in solutions 1&2 to the pH change in solutions 4 & 5. Why is the pH change smaller for solutions 4&5 than in solutions 1&2? Think above the ratio of base and conjugate acid. You can also show this mathematically by using the Kb from part 1 with NH3/NH4* conjugates to determine what the OH- concentration should be. 2
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