9. show work 8. What is the ionic strength of a solution that is 0.0250 M Na:SO,

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9. show work

8. What is the ionic strength of a solution that is 0.0250 M Na:SO, and 0.0150 M Pb(NO,)? (4 pts) a. 0.0800 M b. 0.100 M c. 0.120 M d.0.0600 M e. None of the above 9. Using the extend Debye Huckel equation, what is the activity coefficient for the SO ion from the question 9. The effective radius is 0.40 nm. (4 pts) a. 0.351 b. 0.250 c. 0.327 d. 0.0665 e. None of the above 10. At 25°C, K for the following reaction is 0.040. A mixture of N2, H2, and NHs in a vessel at 25°C has the following concentrations: [N2] 2.0 M, [H.] 0.50 M and [NHs] = 0.10 M. Which one of the following statements concerning the reaction quotient is TRUE for the above system? (4 pts) Note: I made changes to this question that mean the answer on the exam is no longer correct so do not just assume the answer is (d) it is not a. Q K: the system is at equilibrium. b. Q is less than K; more N2 and H2 will be produced. c. Q is less than K; more NH will be produced. d. Q is greater than K; more N; and H2 will be produced. e. Q is greater than K, more NHs will be produced. 11. In the lab we are trying to analyze a solution which contains a triprotic acid of interest. Our spectroscopic method requires the acid to be in the fully deprotonated form for detection. We can control the primary species by altering pH, what pH would you buffer your solution to in order to ensure all of the sample is in the fully deprotonated form? (4 pts) Ka-7.08 x 10". Ka2-1.70 × 10 skal = 6.46 x 10-7 a. pH- 2.00 b. pH 4.00 c. pH 5.00 d. pH-7.00 4I Pag e

Explanation / Answer

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:

[Na+] = 0.025*2 = 0.050, i = 1

[SO4-2] = 0.025; i = 2

[Pb+2] = 0.015; i = 2

[NO3-] = 0.030, i= 1

IS = 1/2(0.050 + 0.025*4 + 0.015*4 + 0.030*1)

IS = 0.12

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)))

= 10^-(0.51*(2^2)*sqrt(0.12) / ( 1 + (400 * sqrt(0.12)/305)))

y = 0.32664

choose c, 0.327

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