The amount of aluminum sulfate required to raise the ionic strength of a 0.1-mol

Question

The amount of aluminum sulfate required to raise the ionic strength of a 0.1-molal solution of calcium nitrate to 0.6 does not exceed 100g. The mean ionic activity coefficient of a solution containing 0 05 mol Ca_3(PI_4)_2/kg H_2O and 0.05 mol FeCl_2/kg H_2O exceeds 1.0. In order for the electrochemical cell potential to reach at least 0.5 V when Zn(s) is oxidized using a 0.1-molal solution of CuCl_2(aq), the molality of the ZnCI_2(aq) product must exceed 0.2. If the gas-solid equilibrium at 273 K and 1 atm amounts to a monolayer capacity of 200 cm^2 on 1.45 g of charcoal using CO_2 (17.0 A^2/molecule), the surface area of this adsorbent does not exceed 800 m^2/g

Explanation / Answer

2) FALSE

Ans = FALSE

Formula to calculate mean ionic activity coefficient (±) is

log± = 0.509z+zI1/2 ( Debye -Huckel limiting law)

where I = ionic strength

Ionic strength I = 1/2 [ no of cations x (cation charge)2 + no of anions x (anion charge)2] x molality

Given that molality of Ca3(PO4)2= 0.05 m

  molality of FeCl2 = 0.05 m

I = I [Ca3(PO4)2] + I [ FeCl2]

= (1/2) [ 3. 22 + 2.32 ] (0.050) + (1/2) [ 1.22+ 2.12] (0.050)

= 0.9

I = 0.9

Calculation of mean ionic activity coefficient :

z+ = 2, z- = 1 z = charge on ion

Ca3(PO4)2 -------------> 3Ca2+ + 2 PO43-

   log± = 0.509z+zI1/2

log± = -0.509 (2x3) (0.9)1/2

log± = -2.89

± = 10-2.89

± = 0.00126

Therefore,

mean ionic activity coefficient = 0.00126 < 1

-------------------------------------

Hence,

ans = FALSE

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