5. You are required to make 1L of each of the buffers below. How much of each in

Question

5. You are required to make 1L of each of the buffers below. How much of each ingredient will you need (mass or volume, as appropriate)? Include calculations with answers Binding Buffer 100mM HEPES (solid) 100mM NaCl (solid or 4M stock solution) 0.05% (v/v) Tween (a liquid of 100% Tween) 2uM DTPA (solid) Remember: Where m- mass (g) n = moles (mol) Mw- molecular weight (g/mol) Dissolution Buffer 8M Urea (solid) 0.1M Tris (1M stock solution) 40mM Glycine (solid) Where n- moles (mol) C concentration (mol/L, M) V - volume (L) Tris Buffered Saline/Tween 25mM Tris (solid) 150mM NaCl (solid or 4M stock solution) 0.1% (v/v) Tween Where Ci-initial concentration Vi-initial volume Cfinal concentration V2- final volume Molecular Weights (g/mol) HEPES: 238.3; NaCl: 58.4; DTPA: 393.4; Urea: 60.7; Glycine: 75.1; Tris: 121.1

Explanation / Answer

Ans. #I. Binding buffer:

# Required moles of HEPES = [HEPES] x Volume of buffer in liters

                                                = 100.0 mM x 1.0 L

                                                = (0.100 mol/ L) x 1.0 L

                                                = 0.100 mol

Required mass of HEPES = Required moles x Molar mass

                                                = 0.100 mol x (238.3 g/ mol)

                                                = 23.83 g

# Required Mass of NaCl = [Conc. x Vol] x MW              ; Conc. in terms of Molarity

                                                = (0.100 M x 1.0 L) x 58.4 g mol-1 = 5.84 g

# 0.05 % (v/ v) means there is 0.05 mL solute per 100.0 mL of solution.

Required volume of Tween = (0.05 mL Tween / 100 mL) x 1000.0 mL = 0.50 mL

# Required Mass of DTPA = [Conc. x Vol] x MW

                                                = (2.0 x 10-6 M x 1.0 L) x 393.4 g mol-1 = 3.934 x 10-4 g

                                                = 39.34 mg

#II. Dissolution Buffer:

# Required Mass of Urea = [Conc. x Vol] x MW

                                                = (8.0 M x 1.0 L) x 60.7 g mol-1 = 485.6 g

# Tris: Using C1V1 (stock soln.) = C2V2 (Final solution)

            Or, 1M x V1 = 0.1 M x 1000.0 mL

            Hence, V1 = (0.1 M x 1000.0 mL) / 1M = 100.0 mL

# Required Mass of Glycine = [Conc. x Vol] x MW

                                                = (0.040 M x 1.0 L) x 75.1 g mol-1 = 3.004 g

# Tris Buffered Saline:

# Required Mass of Tris = [Conc. x Vol] x MW

                                                = (0.025 M x 1.0 L) x 121.1 g mol-1 = 3.0275 g

# Required Mass of NaCl = [Conc. x Vol] x MW             

                                                = (0.150 M x 1.0 L) x 58.4 g mol-1 = 8.76 g

# Tween:        Using C1V1 (stock soln.) = C2V2 (Final solution)

            Or, 100% x V1 = 0.1 % x 1000.0 mL

            Hence, V1 = (0.1 % x 1000.0 mL) / 100% = 1.0 mL

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.