chemistry question on activity

Question

Consider a solution containing the following compounds: 0.01 M NaNO3, 0.002 M Ca(NO3)2 and 0.003 M Al(NO3)3. (i) Calculate the ionic strength of the solution (ii) Calculate the activity coefficients of NO3 –, H+ and OH– in that solution. (iii) Calculate the pH of the solution

Answer 1

First, we'll dissociate all the salts and list the molarity of each ion:

0.01 M NaNO₃ ==> 0.01 M Na⁺ + 0.01 M NO₃^-

0.002 M Ca(NO₃)₂ ==> 0.002 M Ca²⁺ + 0.004 M NO₃^-

0.003 M Al(NO₃)₃ ==> 0.003 M Al³⁺ + 0.009 M NO₃^-

Next, we'll add up the concentrations of NO₃^- ion to obtain final concentration = 0.024 M NO₃^-

(i) Now we can calculate the ionic strength (μ) of the solution using the Debye-Huckel equation. (NOTE: all calculations assume 25ºC):

μ = 1/2 ∑C_iZ_i² where C_i is concentration of ion and Z_i is the charge on the ion

μ = 1/2 (0.01x1 + 0.002x4 + 0.003x9 + 0.024x-1) = 1/2 (0.01 + 0.008 + 0.027 - 0.023) = 0.011 = μ

(ii) Activity coefficients (ϒ) of NO₃^-, H⁺ and OH^-:

log ϒ_z = -0.51z²√μ / 1 + √μ where ϒ_z = activity coefficient

log ϒ_NO3- = -0.51 x 1 x 0.11 / 1 + 0.11μ = -0.0505

ϒ_NO3- = 0.890

ϒ_H+ = 0.913 (obtained from table of activity coefficients when ionic strength is 0.01)

ϒ_OH- = 0.900 (same table as above when ionic strength is 0.01)

(iii) pH:

1x10^-14 = [H⁺]ϒ_H+ x [OH^-]ϒ_OH- = (x)(0.913)(x)(0.900) = 0.8217x²

x = 1.22x10^-7

pH = -log 1.22x10^-7

pH = 6.91

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AFTER SUBMISSION AND REVIEW, I FOUND THE FOLLOWING ERRORS IN MY ORIGINAL ANSWER:

-- Final [NO3-] = 0.023 M not 0.024 M

-- μ = 1/2 (0.01x1 + 0.002x4 + 0.003x9 + 0.023x1) = 1/2 (0.01 + 0.008 + 0.027 - 0.023) = 0.011 = μ (not 0.0105)

I substituted the correct values above. The answer isn't much different as the errors were small