6. (17 points) In this question you are going to calculate the data that could b
ID: 542176 • Letter: 6
Question
6. (17 points) In this question you are going to calculate the data that could be used to construct a titration curve (pH versus volume of titrant added) for the titration of an aqueous ammonia NH3; weak base, pKB = 4.7) cleaning solution with a strong acid (HCI). When strong acid is added to an aqueous solution, it immediately reacts and yields H3O ions, thus the titration is described by the reaction: NH3(aq) _ H,0-(aq) -) NH4-+ H20(1) K = 2.0x109 The cleaning solution is produced by addig 0.02 moles of NH3 to 1.0 L of pure H2O. which vou later titrate with 1.0 M HCl titrant solution. To simplify matters we'll assume all activity coefficients 1.0 for all species in this problem Utilize the ICE Table Method (show your work) to determine the pH of the system when various amounts of the strong acid (titrant) have been added. Note: when you solve the quadratic function, make sure to select the root that is positive, and if you have two positive roots, use the root that represents a realistic change; i.e. the change isn't greater than what was physically added to the system or greater than the concentration of reactant it reacts with. (4 points) Calculate pH of solution when 0 mL of titrant have been added (only the NHs has been added) a. b. (4 points) Calculate pH of solution when 10 mL of titrant have been added c. (4 points) Calculate pH of solution when 15 mL of titrant have been added d. (4 points) Calculate pH of solution when 25 mL of titrant have been added e. (1 points) At what volume of added titrant is the equivalence point reached?Explanation / Answer
when no HCl is added, concentration of NH3= moles/ volume in L=0.02/1 =0.02M
the ionization of NH3 in water is NH3+ H2O------>NH4++OH-, Kb= [NH4+] [OH-]/[NH3]
preparaing the ICE table
component Initial change equilibrium
NH3 0.02 -x 0.02-x
NH4+ 0 x x
[OH-] 0 x x
Kb= 10(-4.7)= 1.995*10-5= x2/(0.02-x), when solved using excel, x= 0.0006219
[OH-] = 0.000629, pOH= -log [OH-]= -log (0.0006219)= 3.21, pH= 14-3.21= 10.79
2. moles of HCl added= 10ml =10/1000L =0.01, moles of HCl added in 10ml =0.01*1=0.01 moles
The reaction between NH3 and HCl is NH3+HCl ---------->NH4Cl, molar ratio of NH3: HCl =1:1 (theoretical)
Actual molar ratio = 0.02 :0.01, so NH3 is excess and moles of NH4Cl formed =0.01
moles of NH3 remainig = 0.02-0.01=0.01, volume after mixing =1+0.01= 1.01
concentrations : NH4+( from NH4Cl) also written as BH+ =0.01/1.01 and [NH3] =0.01/1.01
pOH= pKb+ log [BH+]/[B] =4.7,
pH= 14-4.7= 9.3
3. moles of HCl in 15ml of 1M= 1*15/1000 =0.015, again NH3 is excess, moles of NH4Cl formed =0.015
moles of NH3 remaining = 0.02-0.015=0.005
voume after mixing = 1+0.015= 1.015
concentrations after mixing : NH4+ =0.015/1.015 and NH3=0.005/1.015
pOH= 4.7+ log (0.015/0.005)=5.2
pH= 14-5.2= 8.8
4. when 25 ml of HCl is added, volume of HCl =25/1000=0.025, moles of NH3=0.02
all the NH3 gest consumed and moles of HCl remains
volume of solution after mixing =1+0.025=1.025
concentration of HCl =0.005/1.025= 0.0048
pH= -log (0.0048)= 2.31
at equivalence point , all the 0.02 moles of NH3 is neutralized by moles of HCl
hence moles of HCl =0.02., molarity of HCl =1M
volume of HCl =0.02/1 L =0.02L= 0.02*1000ml= 20ml
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