Can you help calculate the concentrations for 5 and explain to me what effect did NH4+ have on the equilibrium/ph by comparing solutions 2&4. 70°27b/c we areantiloy toHow does this Ka value compare to the value on pg. 1? (within a power of 10 isCalculate ting value what thegood) ThisCalculate the concentrations for solution 4 and solution 5 value is.SolConc. NH3 (M)Conc.NH4+ (M)0.025M20.025MpH10.829.269.2645Compare pH of solutions 2&4. What effect did adding NH4* have on theequilibrium/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? Thinkabove the ratio of base and conjugate acid. You can also show thismathematically by using the Kb from part 1 with NH3/NH4* conjugates todetermine what the OH- concentration should be.2 SOLEREquilibria with Weak Acids & Weak Bases - ResultsExpComposition1A few mL of 0.10M NH3 (unchanged)22.5mL 0.1M NH3 & 7.5mL H₂O32.5mL 0.1M NH4CI & 7.5mL H₂O42.5mL 0.1M NH3 & 2.5mL 0.1M NH4CI & 5mLH₂OTake 1mL Solution #4 and add 9mL of waterGEODOIziprasidons510 mlSolution 1. Use the concentration of NH3 and the measured pH above todetermine the Kb of NH3. (show the ICE table and calculations)-14PH= PKa + log ([A]/[B][AH□ CH]9.26- Pla+log (NHG](NH4)9.24 = Pleat log (a'](0.0Pla = 9,26pkg==logkaKa=-antilog Plaках кв = 10кв = 10-14Ka10-141.81x109Total Voln/a10 ml10 ml10 ml1.81M₂ = (2.5x0.1)/10M2 = 0.025MK6=5,55 x10-24Use the equation (Ka*Kb = 1*10-14) to calculate the Ka for NH4* ?-Antilog (9.26) = -1.81×1091X 10-231pH11.1310.826.019.269.26Solution 2 Use the equation M₁V₁= M₂V₂ to calculate new conc. of NH3.M₁ V₁ = M₂V2Conc. of NH3pH measured: 10.822.5 X0,1 = M2 X10mLm² = 2.5x0.¹ = 0,023MCompare 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 itchange?Kb will not change til the temp changesKb=cok²x²=1,8225x1050.025α = 0,027Solution 3 Conc. of NH4+pH of solution 3: 6.01Use the diluted concentration of NH4+ and measured pH to determine the Ka ofNH4+M₁ V₁ = m₂ V₂61 = 1/2 p²pk2.5ml (0.1M) = M₂ (10m²)[OH-]=CQ=0,025x0,021-log[04²] = -log[0,025 x =0,027)POH= 3,17PH=14-3₁17=108517PH=pKa + log ((A] [B])/([AH] [BH]) ( KaxKb = 1x10"""6.01 = pka +10g ( [0.137 (0.17)Kb = 1x10" / Kapka = 6,01 X10"Ka=-1.02 x 109Kb = 1x10"¹"/1.02X10²Kb = (1/1.02)×10 ²3-23Kb= 0.9804 X16-23K5= 9.804 X10-249How does thgood) ThisCalculate thSol245CompareequilibriuCom

Question

Can you help calculate the concentrations for 5 and explain to me what effect did NH4+ have on the equilibrium/ph by comparing solutions 2&4. 70°27b/c we areantiloy toHow does this Ka value compare to the value on pg. 1? (within a power of 10 isCalculate ting value what thegood) ThisCalculate the concentrations for solution 4 and solution 5 value is.SolConc. NH3 (M)Conc.NH4+ (M)0.025M20.025MpH10.829.269.2645Compare pH of solutions 2&4. What effect did adding NH4* have on theequilibrium/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? Thinkabove the ratio of base and conjugate acid. You can also show thismathematically by using the Kb from part 1 with NH3/NH4* conjugates todetermine what the OH- concentration should be.2 SOLEREquilibria with Weak Acids & Weak Bases - ResultsExpComposition1A few mL of 0.10M NH3 (unchanged)22.5mL 0.1M NH3 & 7.5mL H₂O32.5mL 0.1M NH4CI & 7.5mL H₂O42.5mL 0.1M NH3 & 2.5mL 0.1M NH4CI & 5mLH₂OTake 1mL Solution #4 and add 9mL of waterGEODOIziprasidons510 mlSolution 1. Use the concentration of NH3 and the measured pH above todetermine the Kb of NH3. (show the ICE table and calculations)-14PH= PKa + log ([A]/[B][AH□ CH]9.26- Pla+log (NHG](NH4)9.24 = Pleat log (a'](0.0Pla = 9,26pkg==logkaKa=-antilog Plaках кв = 10кв = 10-14Ka10-141.81x109Total Voln/a10 ml10 ml10 ml1.81M₂ = (2.5x0.1)/10M2 = 0.025MK6=5,55 x10-24Use the equation (Ka*Kb = 1*10-14) to calculate the Ka for NH4* ?-Antilog (9.26) = -1.81×1091X 10-231pH11.1310.826.019.269.26Solution 2 Use the equation M₁V₁= M₂V₂ to calculate new conc. of NH3.M₁ V₁ = M₂V2Conc. of NH3pH measured: 10.822.5 X0,1 = M2 X10mLm² = 2.5x0.¹ = 0,023MCompare 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 itchange?Kb will not change til the temp changesKb=cok²x²=1,8225x1050.025α = 0,027Solution 3 Conc. of NH4+pH of solution 3: 6.01Use the diluted concentration of NH4+ and measured pH to determine the Ka ofNH4+M₁ V₁ = m₂ V₂61 = 1/2 p²pk2.5ml (0.1M) = M₂ (10m²)[OH-]=CQ=0,025x0,021-log[04²] = -log[0,025 x =0,027)POH= 3,17PH=14-3₁17=108517PH=pKa + log ((A] [B])/([AH] [BH]) ( KaxKb = 1x10&#34;&#34;&#34;6.01 = pka +10g ( [0.137 (0.17)Kb = 1x10&#34; / Kapka = 6,01 X10&#34;Ka=-1.02 x 109Kb = 1x10&#34;¹&#34;/1.02X10²Kb = (1/1.02)×10 ²3-23Kb= 0.9804 X16-23K5= 9.804 X10-249How does thgood) ThisCalculate thSol245CompareequilibriuCom

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pH of a solution is calculated as:Buffer: A buffer is a solution that can resist changes in pH when…