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

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


Experiment 9: Determination of the Solubility Product Constant of KHT Reading Chemistry 2.0, OpenStax, Sec. 15,1: Precipitation and Dissolution OBJECTIVE To use the technique of titration to determine the K.o of a solid. INTRODUCTION When solid potassium hydrogen tartrate, KHCiH,OG, (KHT) is added to water, it dissociates into the ions K* and HT (hydrogen tartrate; HCH.O). The structure formula of KHT is: HO OH OH 0 H KHCaH40ss) (KHT) C H -0 K* KHTis the salt of an organic a cid commonly known as cream of tartar by cooks and is used as a stabilizer, thickener, leavening agent (used in baking powder), and pH control agent. It is also formed as a precipitate during the fermentation of grapes during the wine making process. Na Ranes As KHT dissolves in water the concentrations of K* and HT increase and the rates of dissociation (forward reaction) and solid formation (reverse reaction) become equal. Thus, a dynamic equilibrium is reached as represented by the equilibrium reaction in equation 1. At this point the concentration of ions is maximized, i.e., the solution is saturated. K'pg) + HCaH40s (a) (HT) The equilibrium constant expression for this reaction is: Ksp = (K*][HCAH,0s] 9-1 (1) (2) where sp stands for solubility product and the (K'] and [HT] symbolize the molar con centrations (M = mo/L) of K* and HT, respectively. The amount of solid does not affect the equilibrium concentrations as long as some is present. Therefore, concentrations of solids are not included in the equilibrium constant expression; they are assigned values of 1.0. Equation 2 shows that when one mole of KHCaH4Os dissolves, one mole of K* and one mole of HCaH405 are present in the solution. Therefore, the molar ratio of reaction species is 1:1:1 and if the moles of K* or HCHe0s in a saturated solution can be determined then the moles of KHT dissolved can be determined since they wil be the same for this reaction. You will determine the number of moles of HT present in a saturated solution of KHT using the titration technique. HT (a monoprotic acid) will be titrated with NaOH via the net-ionic reaction (3) Because one mole of sodiun hydroxide neutralizes one mole of HT ions, the number of moles of sodium hydroxide used in the titration will equal the number of moles of HT. The (HT] will be equal Two Videos What is K,? (Solubility Product Constant) & Solubility Equilibria Concentration of standard NaOH Volume of saturated KHT solution Temperature of KHT solution Initial buret reading Final buret reading Volume of NaOH delivered Moles of NaOH used Moles of HT titrated Show one calculation in the space below. Trial I Tnal a Trial 3 Trial 1 Show one calculation in the space below. (HT] in KHT(ag) in each trial 50m 0.03 (K] in KHT(ag) in each trial Show one calculation in the space below. Show one calculation in the space below. Trial 2 O.03 9-3 Trial 3 5Uml Ksp Show one calculation in the space below. Average Ksp Show calculation in the space below. In this experiment, a volume of NaOH is used to titrate a solution of KHT. Explain how it is possible to use the moles of NaOH required in the titration to determine the Ksp of KHT. 9-4 Additional Questions 1) Write the equilibrium reaction for the dissociation of solid copper(!) sulfide, Cu;S in water. 2) Write the equilibrium constant expression, K,o, for the CuzS reaction. 3) Applying Le Chatelier's Principle., how will the addition of the following substances affect the equilibium reaction written in #1 and how will the equilibrium be restored? Explain. In completing your answer, complete the following equations to show how these compounds exist when in water. a) NazS(s) b) CuCl(s) c) NaCl(s) water e) CuzS(s) Water Water d) Temperature is increased (reaction is endothermic) 4) Using the information from 1 and 2 above, calculate the concentrations of [Cu'] and (S?-] if the Kso of CuS(s) is 6.0 x10-", Hint: Assume (Cut] = 2x and [S]=x and then solve for x after setting up the equilibrium constant expression. 9-5 Name: Experiment 9: Determination of the Solublity Product Constant of KHT Na kainos Quiz Practice Questions 1. Consider solid aluminum carbonate, Al;(Co;)1(s). Instructor: (a) Write a balanced equilibrium reaction equation for the dissociation of the compound. Sec: (b) Write an equilibrium constant expression (the solubility product constant, Kso) for the for the aqueous dissociation of the compound. 2. Write the equilibrium reaction and solubility product constant (Ksp) for the dissolution of Hg,Cl,(s). Caiculate the Ksp Value if the experimentally determined equilibrium concentrations are: (Hg,*] = 1.78 x 10 M and [CI= 3.97 x 10 M, respectively. 9-6 3. Write the equilibrium reaction and solubility product constant (Ksp) for the dissolution of solid Zna( PO.)2. Calculate the zinc ion and phosphate ion concentrations. the Ksp Of Zn,(PO.), is 9.1 x10-33
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
34 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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