MCB 2000: lab practical Study Guide Exam with All Correct Answers Updated.

MCB 2000: lab practical Study Guide
Exam with All Correct Answers 2025-
2026 Updated.
Starting with a 0.5 M solution of NaCl, how would you make 500mL of a 150 mM solution of
NaCl? Clearly explain. - Answer You would add 150mL of the 0.5M NaCl solution and then
add 350mL of water to make 500mL of a 150mM solution.



Math:



(500mL)(0.15M) = (0.5M) V

V = 150 mL solution



500mL total - 150 mL solution = 350mL water



Convert:

150 μL to mL

0.075 M to mM

0.950 mL into μL - Answer 1. 0.15 mL

2. 75 mM

3. 950uL



The pH of a solution with [H+] concentration less than water would be... - Answer greater
than 7



HEPES has a pKa of 7.55. At what pH would a HEPES buffer have the maximum buffering
capacity? - Answer 7.55



How many pKa values would you predict for the amino acid aspartic acid? - Answer 3



Which of the following statements below is NOT correct regarding thin layer chromatography:

, 3. Using pen, write your initials and section # on the plate



4. Samples must be spotted on the side of the plate coated with silica (not the shiny plastic side)
- Answer 3. Using pen, write your initials and section # on the plate



T/F:

In the Buffers and pH lab, you will be using two techniques to identify an unknown amino acid:
thin layer chromatography and titration - Answer true



Starting with a 0.5 M solution of NaCl and a 0.1 M solution of NaH2PO4, how would you make
500 mL of a solution that contains 150 mM NaCl and 20 mM NaH2PO4? - Answer To make
this solution you would mix 150 mL of the 0.5 M solution of NaCl, 100 mL of the solution of 0.1
M NaH2PO4, and 250 mL of water.



math:



NaCl:

C1= 0.5M

V1= ?

C2 = 0.15M

V2= 500mL



(0.5M)V2 = (0.15M)(500mL)

V2 = 150 mL NaCl



NaH2PO4:

C1= 0.1M

V1= ?

C2 = 0.02M

V2= 500mL



(0.1M)V2 = (0.02M)(500mL)


V2 = 100 mL NaH2PO4