How many gramms of Tris and Tris Hcl are needed to make up 500ml of 1m Tris with a PH of 8.0? (pka of Tris is 8.3, mw of Tris is 121, mw of Hcl is 157.
Pka is Tn3 =8.3
MW of Tn3 =121
MW of Tn3 HCL=157
From the question given: Tris Hcl us the Acid,with Mw=157.and then Tris is the base with the Mw=121. then converts 500mls to litre which is 0.5 litre then from the Henderson and Hasselbach equation PH=pka+log(A/H) then 8.0-8.3=log Tri/Tri Hcl then -0.3=log(Tri/Trid HCl =0.50 from antilog of -0.3= Tris/TrisHCl ......... Loading
To determine the amount of Tris (Tris base) and Tris HCl (Tris hydrochloride) required to make up 500ml of 1M Tris with a pH of 8.0, we need to consider the pKa of Tris and the desired pH.
First, let's understand the relationship between pH and pKa. The pKa is the pH at which a weak acid is 50% dissociated into its conjugate base. In this case, Tris acts as a weak base. Since the Tris pKa is 8.3, this means that at pH 8.3, half of the Tris will be present as Tris base and half as Tris HCl.
To achieve a pH of 8.0, which is slightly lower than the pKa, we need a higher concentration of Tris base compared to Tris HCl. Considering only the Tris base, we can calculate the required concentration:
pH = pKa + log(base/acid)
8.0 = 8.3 + log(base/acid)
Rearranging the equation, we can solve for the ratio of base to acid:
log(base/acid) = 8.0 - 8.3
log(base/acid) = -0.3
Next, we need to convert this ratio into the actual molar quantities. The molecular weight (MW) of Tris base is given as 121 g/mol, and the MW of Tris HCl is 157 g/mol.
Let's assume the amount of Tris base is x grams.
Therefore, the amount of Tris HCl is x grams as well.
Using the equation:
(base/acid) = 10^(log(base/acid))
(base/acid) = 10^(-0.3)
(base/acid) = 0.5012
We can now set up a ratio using the molecular weights:
(x g Tris base / 121 g/mol) = (x g Tris HCl / 157 g/mol)
Solving for x:
0.5012x / 121 = x / 157
Cross-multiplying and solving the equation, we find:
0.5012 × 157 = 121 × x
78.6854 = 121 × x
x = 78.6854 / 121
x ≈ 0.65 g
So, approximately 0.65 grams of both Tris base and Tris HCl are needed to make up 500ml of 1M Tris with a pH of 8.0.