Question

Assume each salt disassociates 100% a) Calculate the ionic strength of a 500mM solution of NaCl b) Calculate the ionic strength of a 500mM solution of CaCl2 c) Discuss your answers to a and b. Given they are the same concentration are the ionic strengths equal or different? Explain.

Answer #1

a)

NaCl --> Na+ and Cl-

charges are +1 and -1

[Na+] = 500 mM = 500*10^-3 M = 0.5 M

[Cl-] = 500 mM = 500*10^-3 M = 0.5 M

Recall that ionic strength considers all ions in solution, and its charges. It is typically used to calculate the ionic activity of other ions. The stronger the electrolytes, the more ionic strength they will have.

The formula:

I.S. = 1/2*sum( Ci * Zi^2)

Where

I.S. = ionic strength, M (also miu / μ ) used

Ci = concentration of ion “i”

Zi = Charge of ion “i”

The exercise:

I.S. = ionic strength, M (also miu / μ ) used

I.S = 1/2* ((0.5)*(1^2) + (0.5)(-1)^2))

I.s = 0.5 M

b)

CaCl2

Ca+2 and 2 Cl- ions

[Ca+2] = 0.5 M

[Cl-] = 2*0.5M = 1 M

I.S = 1/2* ((0.5)(2^2) + (1)(-1)^2)

I.S = 1.5 M

c)

the ionic strenght is "felt" stornger in CaCl2 due to:

- Charges (+2, -1)

- amount of ions present, i.e. 3 ions

These will appear to have a higher interaciton of ions in solution

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