Question

Starting with two stock buffer solutions (A and B) listed below, describe how you would mix them together to obtain the solutions i and ii. Write your answers to 3 significant figures. i) 400 mL of 0.35 M NaCl, 10 mM Tris buffer (pH 8.0) ii)600 mL of 0.9 NaCl, 10 mM Tris buffer (pH 8.0). Stock buffer A: 3 M NaCl, 10 mM Tris buffer (pH 8.0) and stock buffer B: 0.1 M NaCl, 10 mM Tris buffer (pH 8.0). Please show work! Thank you.

Answer 1

¡) 400ml of 0.35M NaCl , 10mM Tris buffer

X mol + (0.21 - X mol ) = 0.14mol

((3mol/1000ml)×Xml) + (0.1mol/1000ml)×(400 -X)ml = 0.14mol

0.003X + 0.040 - 0.0001X = 0.14

0.0029X = 0.10

X =34.48ml

400 - X = 365.52ml

So,

34.48ml of stock A mixed with 365.528ml of Stock B

¡¡) 600ml of 0.9M NaCl, 10mM Tris

( (3mol/1000ml)×Xml ) + ( 0.1mol/1000ml)×(600-X)ml = 0.54mol

0.0030X + 0.060 - 0.0001X = 0.54

0.0029X =0.48

X = 165.52ml

600 - X = 434.48ml

Therefore,

165.52ml of Stock A mixed with 434.48ml of stock B