- #1

jibjab

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## Homework Statement

To purify osprase you must use 750 mL of an extraction buffer consisting of 0.500 M Tris, pH 7.8, 100 mM NaCl,

20% sucrose. You have the following chemicals available:

Tris base (FW = 121.1 g/mol; pKa = 8.1)

1.00 M HCl

NaCl (FW = 58.4 g/mol)

Sucrose (FW = 342 g/mol)

Outline how you would prepare your solution, show all calculations.

## Homework Equations

pH=pKa+log(B)/(A)

## The Attempt at a Solution

I plugged in the pH and pKa in the equation and got 0.5012A=B.

I multiplied 0.75L x 0.5 mol/L = 0.375 mol

I solved for A = 0.2498 mol then multiplied that by 1000mL/1mol = 249.8 mL HCl

I solved for B = 0.1252 mol x 121.1 g/mol = 15.16 g Tris base

^The biggest thing I need to know is how do you know what to do with your B value? What I mean is how do you know when you have to multiply the formula weight of the tris base with the B+A value(0.375 mol in this case), or when you should solve your A value, subtract the B+A value from the A value to solve for B, then multiply the FW with the B value for the amount of base added?

(Feeling around in the dark for this section)

I multiplied 0.75 L x .01 mol/L = 0.075 mol NaCl

0.075 mol NaCl x 58.4 g/mol = 4.38 g NaCl

0.375 mol x 342 g/mol Sucrose = 128.25g x 0.2 = 25.65 g Sucrose.

^I haven't solved one that had other stuff in the buffer before, I'm not sure what to do with that bit.

Any explanations provided are highly appreciated! :)