Question

How many ul of a 350 mM CaCl2 solution would you need to pipette to dispense 25 umoles of Cl- ions?

u=micro

Answer 1

Consider the dissociation of CaCl₂.

CaCl₂ -------> Ca²⁺ + 2 Cl^-

As per the stoichiometric equation,

1 mole CaCl₂ = 2 moles Cl^-

Therefore, 25 µmol Cl^- = (25 µmol)*(1 mol CaCl₂/2 mol Cl^-) = 12.5 µmol CaCl₂.

Let the volume of 350 mM CaCl₂ required be V L. Plug in values and get

V = (12.5 µmol CaCl₂)/(350 mM CaCl₂) = (12.5 µmol)*(1 mol/10⁶ µmol)/[(350 mmol/L)*(1 mol/1000 mmol)] = (1.25*10^-5 mol)/(0.35 mol/L) = 3.5714*10^-5 L = (3.5714*10^-5 L)*(1.0*10⁶ µL/1 L) = 35.714 µL ≈ 35.7 µL (ans).