MasteringChemistry Homework: Chemical Equilibria (Chapters 17 amp: 18 Google Chr
ID: 1025912 • Letter: M
Question
MasteringChemistry Homework: Chemical Equilibria (Chapters 17 amp: 18 Google Chrome secure https://session.masteringchemistry.com/myct/itemView?assignmentProblemID=95741888 The butter trom Part A Is dilluted to 1.00 L. To halt of It (500. mL), you add 0.0100 mol of hydrogen lons without changing the wolume. What Is the pH or the final solution? In the study of biochemical processes, a common burfering agent is the weak base trishydroxymethylaminomethane (HOCH2CNH2, orten abbreviated as Tris. At 25 °C, Tris has a pKh of 5.91 Express your answer numerically to two decimal places. View Available Hint(s) pH 7.58 The hydrochloride of Tris is (HOCH2 CNH Cl, which carn be abbrevlated as TrisHCL. Previous Answers Correct Notice that even though a significant amount of the strang acid was added to the buffer, the pH did nat change drasticaly since the capacity of the buffer had not been excecded. Significant Figures Feedback: Your answer 7.57 was either rounded differently or used a different number of significant figures than required for this par Part C What additional volume of 10.0 M IICl would be needed to exhaust the remaining capacity of the buffer after the reaction described in Part B? In other words, how much more of this HCl sclution is required to consume the remaining Tris in the buffer? Express your answer in milliliters using two significant figures. View Avallable Hint s) SubmitExplanation / Answer
pH = 7.79 , pOH = 14-pH = 14-7.79 = 6.21
pOH = pkb + log [TrisHCl]/[Tris] ,
TRIHCl moles = 31.52/157.64 = 0.2 moles
( 157.64 = mol wt of TRISHCl)
TRis HCl + NaOH ---------> TrIs + H2O
6.21 = 5.91 + log [TRis HCl]/[Tris]
[Tris HCl] = 1.9953 [Tris]
Tris HCl moles = 1.9953 Tris moles ( we know by eq Tris moles = NaOH moles = 10 V , V= vol)
( 0.2 - 10 x V) = 1.9953 ( 10V)
V = 0.00667 liters = 6.67 ml
b)
vol = 1 liter
half solution is taken , hence using M1V1/n1 = M2V2/n2 here M is same but half vol is taken ,
hence moles become half
V2/V1 = n2/n1 = 500/1000 = 0.5 , n2 = 0.5n1
hence Tris HCl moles = (0.2-10x0.00667)/2 = 0.06665
Tris moles = ( 10x0.00667)/2 =0.03335
H+ added = 0.01
hence net Tris HCl = 0.06665+0.01 = 0.07665
net Tris = 0.03335-0.01 = 0.02335
pOH = pka + log [Tris HCl]/[Tris]
pOH = 5.91 + log ( 0.07665/0.02335) = 6.426
pH = 14-6.426 = 7.574