In the study of biochemical processes, a common buffering agent is the weak base
ID: 793309 • Letter: I
Question
In the study of biochemical processes, a common buffering agent is the weak base trishydroxymethylaminomethane,
The hydrochloride of Tris is (HOCH2)3CNH3Cl, which can be abbreviated as TrisHCl.
Question 1.) What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl?
Question 2.) The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0100mol of hydrogen ions without changing the volume. What is the pH of the final solution?
Explanation / 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
2) 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