Question

21 Calculate the ionic strength and the mean activity coefficient of 1.00 mmol kg CaCl, (aq) at 25°C. Calculate the masses of (a) KNO, and, separately, (b) Ba(NO3), to add to a 0.110 mol kg solution of KNO, (aq) containing 500 g of solvent to sitionic strength to 100. b

Answer 1

21 (a)

Ionic strength (I) = 1 ² ci zi ci concentration of lon t charge on the iont For cach, Z Z + 2 Given that 1.00 mmol leg! cach 7 100 x10 mol kg cache 0.001 mol kg cach 0,ool nolal solution of cach 2+ Ca cach + 2 cl 0.00TK 2 0.00lm 0.00 lmates = 0.00 2. 너 0. Dolm Concentration of a conx Cat concentration of ct ion ci 0,00 2m charge on cation +2 - charge on anion Ecí [..c01* 6++ 0.602 XC-1)] } [o.col x4 + 0,00 2x]]

=> LE - 12.004 + 0.002] 1 [0.006) 2 1 0.003 Tonic strenth of a Innal kat of cach, sol? is 0.003 Accordig to Debye tuckel Limiting Law - AIZZ 15E log de here - mean activity co-efficient oro 0.509 at 25°c Debye -Iteckel constant +2 Z charge on cation = change on anion strength Ionic 0.003 1 i log & 0. 5.9 x 1 +2 -11 x Jo.003 - O 509 X1-21 X 0.0547722558 = -0.509 * 2 *0.05477 22558 logit = -0.0557581564 Take anti logarithms on both sides logt clog &7) Tog' c-0.0557 58 1564) 0.8795122 3 B+ À - 0.8795

(b)

We'll start with the definition of ionic strength (eq 5.76 on pg 196) I= z? (6/6) For part (a) we have two ions, K+ and NO3, with charges of 2 = +1 and 2 = -1 respectively. I = z? (6/6) } [(+1)?bk+/6Ⓡ (-1)ºbxo;/B] b/be with b as the concentration of KNO3; b = bk+ = bnoz

We can construct an expression for b in terms of the initial concentration by and the moles n of KNO3 added to the solution volume Msolv n b=bo + Mgolv Additionally, as we're interested in mass of KNO, we can replace moles n with mass m using molecular weight MW,KNO3- m b=bo + Msoly Mw,KNO3 This can be substituted back into our expression for ionic strength I and the resulting equation can be solved for m (mass of KNO3 to add). m (16e – bo) Msoly Mw,KNO3 (1.00 mot kg? -0.110 mot kg 44.994 g x 0.500 kg x 101.11 gmolt For part (b) we'll once again begin with the definition of ionic strength and this time we'll include the ion Ba+2 with 2= +2. FIA I bBa(NO3)2 + [(+1) bBa(NO3)2 +(+2) +(-1) be DKNO3 DKNO3 +(-1) be be 3bBa(NO3)2 be BKNO be +

where in this equation bKNO, is the initial and constant) concentration of potassium nitrate and bBa(NO3), is the concentration of barium dinitrate created by adding Ba(NO3)2 to the solution. Similarly to part (a) we'll construct an expression for bBa(NO3), in terms of mass of Ba(NO3)2 added m. m bBa(NO3)2 Msoly Mw,Ba(NO3)2 Substituting this into our expression for I and solving for m gives mn (Ibe – BKNO,) Msoly Mw,KNO, 3 (1.00 mot kg - 0.110.mot ke-1) x 0.500 kg x 261.32 g.mot+ 3 38.762 g Solution • (a) m= 45.0 g • (b) m = = 38.8g

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