(a) Calculate the ionic strength of a solution of 0.2 M Na2SO4 using Equation 1,

Question

(a) Calculate the ionic strength of a solution of 0.2 M Na2SO4 using Equation 1, assuming that there is no ion-association between Na+ and SO4 2- . Equation 1: I = 0.5(cizi 2 ) (b) Predict whether the ionic strength of 0.2 M Na2SO4 would be higher or lower than the value calculated in part (a) if the association Na+ + SO4 2- = NaSO4 - was taken into account? Give reasons for your answer. (c) Calculate the concentration of NH3 in a solution of 1 M total ammonia maintained at pH 10.2 and 298 K using Equation 2. Equation 2: pH = 9.2 + log {[NH3]/[NH4 + ]} (at 298 K) (d) Calculate the concentration of Ag+ ions in water saturated with AgCl at 298 K using Equation 3 and express your answer in mg/L. Equation 3: AgCl(s) = Ag+ + Cl- Ks = 10-9.7 at 298 K (e) Calculate the concentration of AgCl3 2- in a solution of 3 M MgCl2 saturated with AgCl at 298 K using Equations 3 and 4 and express your answer in mg/L. Equation 4: Ag+ + 3Cl- = AgCl3 2- 3 = 106.2 at 298 K

Explanation / Answer

a) The dissociation equation is

Na2SO4 (aq) <====> 2 Na+ (aq) + SO42-

The molarity of the ions and the charges are:

Na+ : (0.2*2) M = 0.4 M; +1

SO42-: (0.2*1) M = 0.2 M; -2

The ionic strength of the solution is

I = 0.5[0.4*(+1)2 + 0.2*(-2)2] = 0.5[0.4*1 + 0.2*4] = 0.5[0.4 + 0.8]

= 0.5*1.2 = 0.6 (ANS)

b) When we consider the association, we are considering that a positively charged Na+ is surrounded by negatively charged SO42-. There is some interaction of charges between the oppositely charged ions. This reduces the formal charge on an ion to a value less than the normal value. Since the formal charge decreases, the ionic strength of the solution must decrease.

c) The total ammonia concentration is maintained at 1.0 M; this means

[NH3] + [NH4+] = 1.0 M ……(1)

Also, since the pH of the solution and the pKa values are given, we must consider the dissociation reaction

NH4+ (aq) <====> NH3 (aq) + H+ (aq)

The Henderson-Hasslebach equation for the solution is

pH = pKa + log10[NH3]/[NH4+]

Therefore, 10.2 = 9.2 + log10[NH3]/[NH4+]

===> 1.0 = log10[NH3]/[NH4+]

===> [NH3]/[NH4+] = 101 = 10

===> [NH3] = 10[NH4+] …..(2)

We have two equations and two unknowns. Combining (1) and (2), we have

10[NH4+] + [NH4+] = 1.0 M

===> 11[NH4+] = 1.0 M

===> [NH4+] = 1.0 M/11 = 0.091 M

Therefore, [NH3] = 1.0 M – 0.091 M = 0.909 M

Ans: The equilibrium concentrations are: [NH3] = 0.909 M and [NH4+] = 0.091 M

d) The dissociation equation is

AgCl (s) <====> Ag+ (aq) + Cl- (aq)

Ks = 10-9.7 = 1.995*10-10 2.0*10-10

Also, Ks = [Ag+][Cl-] =[Ag+]2 (since [Ag+]=[Cl-]=molar solubility of AgCl)

===> 2.0*10-10 = [Ag+]2

===> [Ag+] = 1.414*10-5

The molar solubility of AgCl is 1.414*10-5 M

Next we convert molar solubility to gm/L bu multiplying by the molar mass of AgCl = 142.32 gm/mol.

Molar solubility = 1.414*10-5 mol/L * 142.32 gm/mol = 2.013*10-3 gm/L

Divide by 1000 to get molar solubility in gm/mL = 2.013*10-6 gm/mL = 2.013 mg/mL (ans)

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