posted by Rose
How do you prepare 100 mL of a 0.1 M TRIS-HCl buffer with a pH 7.4 using solid TRIS base (MW 121.1) and 1 M hydrochloric acid?
I started with
pH=pKa + log ([Tris]/[Tris+]
so 0.219:1 ratio with 2.219 total
[Tris}=(0.219/1.219)*0.1 M= 0.018 M
[Tris+]= (1/1.219)*0.1 M= 0.082 M
M_[Tris]= (0.018M)(0.1L)= (0.0018 mol* 121.1 g/mol)= 0.218 g Tris
I'm not sure if this is correct but my main issues is how do I determine the amount of HCl I will need to add?
Since you are starting with Tris and you must generate the Tris(HCl, I would start with 0.1L x 0.1 M Tris or 0.01 moles Tris which is 1.211 grams. Now you want to add enough 1M HCl to make Tris*HCl (the acid) but keep the ratio of base/acid right for it to end up a pH of 7.4
................Tris + HCl ==> Tris*HCl
7.4 = 8.06 + log [(0.01-x)/(x)]
x = 0.0082 moles HCl (which is 8.2 mL of 1M HCl). You need to confirm that.
1.211 g Tris = 1.211/121.1 = 0.01 moles (or 10 millimoles).
Add 8.2 mL of 1 M HCl is 8.2 millimoles HCl added.
That will form 8.2 mmoles of the Tris*HCl and will leave 10 mmoles base-8.2 mmoles HCl = 1.8 mmoles free base.
pH = 8.06 + log(1.8/8.2) = 7.401 which looks close enough to me and rounds to 7.40.
How do I calculate new pH of the buffer if I were to add 1mL of 1 M HCl to 5mL of buffer?
Where did you get a pKa of 8.06?
I think the 8.06 came from 14-pKb of Tris (5.924)