You are trying to model a system based on seawater, so you create a solution tha

Question

You are trying to model a system based on seawater, so you create a solution that contains the approximate concentrations of the major ionic components of seawater: 0.450 M NaCl, 0.025 M MgCl2, 0.025 M MgSO4, and 0.010 M CaCl2.

a. What is the ionic strength of the solution?

b. What method would best to determine the activity coefficient for hydronium ions in a solution with this ionic strength?

c. Suppose you took a 1/10 dilution of this solution, while adding enough hydrochloric acid to reach 0.0125 M HCl in the final solution. Taking into account activity, what is the expected pH of the final solution?

Explanation / Answer

a)

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 / ) used

Ci = concentration of ion “i”

Zi = Charge of ion “i”

The exercise:

IS = 1/2*( (0.45)(1) + (0.45)(1) + (0.025)(2^2) + (0.050)(1) + (0.025)(2)^2 + (0.025)(1) + (0.01)(2^2) + (0.02)(1))

IS = 0.6175

b)

the best ways iv via Deby Huckl equation

Activity of X = x * [X]

Where:

Activity coefficient () of “x”

[X] = molar concentration concentration of X

Note that Activity coefficient () depends on

Where

i = activity coefficient for species “i”

i = theoretical diameter in pm (10^-12 m)

Zi = Charge of ion

I.S. = ionic Strength (usually used as as well)

If we wanted only

= 10^-(0.51*(Zi^2)*sqrt(I.S.) / ( 1 + ( * sqrt(I.S)/305)))

Zi = +1 for H++, alpa = 900; IS = 0.6175

= 10^-(0.51*(Zi^2)*sqrt(I.S.) / ( 1 + ( * sqrt(I.S)/305)))

= 10^-(0.51*(1^2)*sqrt(0.6175) / ( 1 + (900* sqrt(0.6175)/305)))

-H+ = 0.7572

c)

recalculate ionic strenght; al species are 0.1 diluted so:

IS = 1/2*( (0.045)(1) + (0.045)(1) + (0.0025)(2^2) + (0.0050)(1) + (0.0025)(2)^2 + (0.0025)(1) + (0.001)(2^2) + (0.002)(1))

IS = 0.06175

re-substitute in activity of pH:

= 10^-(0.51*(1^2)*sqrt(0.06175) / ( 1 + (900* sqrt(0.06175)/305)))

diluted= 0.8450

pH = -log(ACtivity of H) = -log(*[H+])

pH = -log(0.8450*(10^-7))

pH = 7.07

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