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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+]
[Tris]/[Tris+]= 10^(7.4-8.06]=0.219
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?

  • Biochemistry -

    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
    initial mols...0.010....0........0
    final.........0.010-x...x....... x

    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.

    Check 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.

  • Biochemistry -

    How do I calculate new pH of the buffer if I were to add 1mL of 1 M HCl to 5mL of buffer?

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