I have a lab due on Wednesday before 5 am I need some assistance.

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Answered 1 days AfterApr 10, 2023

Answer To: I have a lab due on Wednesday before 5 am I need some assistance.

Baljit answered on Apr 12 2023
37 Votes
Experiment 9
Determination of the Solubility Product Constant of KHT
Lab report
OBJECTIVE
· To use the techni
que of titration to determine the Ksp of a solid
Calculations:-
Concentration of standard NaOH, MNaOH = 0.05M
    
    Formula
    Trial 1
    Trial 2
    Trial 3
    Volume of saturated KHT solution
    
    50ml
    50ml
    50ml
    Temperature of KHT solution
    
    56.6oC
    56.6oC
    56.6oC
    Initial buret reading
    
    0.02
    0.02
    0.02
    Final buret reading
    
    6.5ml
    14ml
    19.6ml
    Volume of NaOH delivered (VNaOH)
    
    48ml
    49ml
    50ml
    Moles of NaOH used
    =VNaOH(liters)*MNaOH
    =0.05*0.048
=0.0024mol
    =0.00245 mol
    =0.0025 mol
    Moles of HT titrated
    =Moles of NaOH used
    0.0024mol
    0.00245 mol
    0.0025 mol
    (HT-] in KHT(ag) in each trial
    =
    =0.0024/0.05
=0.048M
    0.049M
    0.050M
    (K] in KHT(ag) in each trial
    
    0.048M
    0.049M
    0.050M
    Ksp
    =
    =0.0023040
    =0.002401
    =0.0025
· Average Ksp=(0.002304+0.002401+0.0025)/3=0.002402
· In the following reaction
Since One mole of NaOH is required to neutralize one mole of HT- so number of moles of NaOH used in titration is equal to number of moles of...
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