Ch.7 ACTIVITY & SYSTEMATIC TREATMENT OF EQ

Transcription

1 Ch.7 ACTIVITY & SYSTEMATIC TREATMENT OF EQ 7. Activity Effective concentration under ionic surroundings 7-. The effect of Ionic Strength on Soluility of salts. * Consider a saturated sol. Hg (IO 3 ), concentration of Hg +? Hg (IO 3 ) Hg + + IO 3 - sp =.3 x 0-8 sp = [Hg + [IO 3- = x(x) x = [Hg + = /[IO 3- = 6.9 x 0-7 M this is the max. [Hg + that can e dissolved. only when they are in distilled water without any other ions. * If M NO 3 is added to the sol, ( + does not react with IO 3- ) 7-. The effect of Ionic Strength on Soluility of salts. 7. more solid (Hg (IO 3 ) ) dissolves until = [Hg + =.0 x 0-6 M (50% increase) Addition of "inert" salt (NO 3 ) to insolule salt (Hg (IO 3 ) ) Increase in soluility Why? With more ions, ionic atmosphere uilds up net attraction etween cation - anion ecomes smaller than pure cation and anion. (atmospheric counter charge decreases the ionic charge) increasing the soluility

2 The effect of Ionic Strength on Soluility of salts. Increasing ionic strength promotes the dissociation of compounds into ions MX M + + X - decrease in product conc. y eing attracted to the previously dissolved ions : Le Chatelier's principle MX dissolves more What is Ionic Strength? : measure the total conc. of ions in solutions. I (c z c z... ) c i z i c i : concentration of ith species z i : charge (+, +) 7- Activity Coefficient 7.4 For expressing the concentration influenced y other ionic presence, Activity must e used. Activity of C : A c = [C c c : activity coeff. a A + B c C + d D Real eq. const. : A A c C a A A A d D B Thermodynamic eq. const. : Activity Coefficient [C [A c a c C a A [C [A [D [B c a d [D [B d D B d ionic strength at low ionic strength < at high ionic strength

3 Activity Coefficient of Ions 7.5 Deye Huckel's definition 0. 5z log ( 0.M ) at 5 0 C ( / 305) : ion size parameter : effective hydrated radius small, highly charged ions inds solvent molecule more tightly, and have larger hydrated radii. F I Using Activity Coefficient 7.6. A Simple Soluility prolem * Concentration of Ca + saturated with CaF in a 0.05 M soln. of MgSO 4 CaF (s) Ca + + F - sp = 3.9 x 0- initial: 0 0 final: x x ) without MgSO 4 sp = x(x) = 3.9 x 0 - x =.36 x 0-4 M = [Ca + ) with 0.05 M MgSO 4 A A [Ca [F sp Ca x(x) = /(0.05x x(-) = M x = 3. x 0-4 M = [Ca + 45% increase in soluility Ca F Ca , 0. 8 F F Ca F

4 Using Activity Coefficient 7.7. The Common Ion Effect In the case of common ion existed [Ca + of CaF with M NaF (I=0.05), [F - : common ion CaF Ca + + F - initial final x x+0.05 If x is very small, x << Since I=0.05, is the same as aove. sp [Ca Ca [F x(0.485)(0.50) F (0.8) 3.9x0 x = [Ca + = 4.9 x 0-8 M Decrease in soluility due to common ion which affects the total product sp 7.8

5 ph Revisited (from the point of Activity) 7.0 ph = - log [H + ph loga log[h H if we use ph meter, it sees actual hydrogen ions only. ex) ph of water containing 0.0 M Cl at 5 0 C. Cl H 0. 0M 0. 83, OH w = x(0.83)x(0.76) =.0 x 0-4 x =.6 x 0-7 M ph = 6.98 H

6 7-4. SYSTEMATIC TREATMENT OF EQUILIBRIUM 7. Understand the chemical equilirium Basis for most analytical techniques.. Charge Balance Electroneutrality : solutions must have zero total charge z + = z - Mass Balance : Conservation of Matter 7-4. SYSTEMATIC TREATMENT OF EQUILIBRIUM 7. A solution y dissolving mol acetic acid in Liter, it partially dissociates to CH 3 COOH CH 3 COO - + H + acetic acid acetate M = [CH 3 COOH + [CH 3 COO - what we put undissociated dissociated in the solution

7 7-4. SYSTEMATIC TREATMENT OF EQUILIBRIUM 7.3 *In case of mol/l H 3 PO M = [H 3 PO 4 + [H PO 4- + [HPO [PO 4-3 * For La(IO 3 ) 3 in water La(IO 3 ) 3 La IO 3 - [IO 3- = 3 [La Applying the Systematic Treatment of Eq. 7.4 step : step : step 3: step 4: step 5: step 6: write all reactions write charge alance write mass alance write equilirium constant count the equations & unknowns solve for unknowns

8 7-5 Applying the Systematic Treatment of Eq Ionization of water (autoionization) step : H O H + + OH - w =.0 x 0-4 at 5 0 C -3: [H + = - 4: w = [H + - H OH =.0 x 0-4 for pure water, ph = - log A H+ = 7.0 H OH We want to omit activity coefficient for simpler calculations. Let's use concentration for the later chapters. 7-5 Applying the Systematic Treatment of Eq A solution of Ammonia mol NH 3 in.000 L. To find [NH 3, [NH 4+, [H +, - step : charge alance : sum of positive = sum of negative Step 3: mass alance Step 4: Eq expressions

9 7.7 Without counting activity effect, Soluility of Calcium Sulfate step : charge alance : sum of positive = sum of negative Step 3: mass alance

10 Soluility of Manganese Hydroxide step : charge alance : sum of positive = sum of negative Step 3: mass alance 7.0 Simplification: in asic solution, - >> [H + neglect [H + [MgOH + = [Mg + =

11 7-5 Applying the Systematic Treatment of Eq. 7. Case of the product of one reaction is a reactant in the next reaction. 4. Soluility of CaF CaF (s) Ca + + F - sp = 3.9 x 0 - only F - can react with water to give HF (aq) (Assume no further reaction of Ca + ) F - + H O HF + OH - =.5 x 0 - In any case, water should meet the following condition. (H O H + + OH - w =.0 x 0-4 ) 7-5 Applying the Systematic Treatment of Eq. 7. Systematic Treatment step : charge alance : sum of positive = sum of negative [H + + [Ca + = - + [F - step 3: [F - =[Ca + at first eq. [F - total = [F - remained + [HF at final step 4: sp = [Ca + [F - = 3.9 x 0 - [HF. 5x0 [F w = [H + - =.0 x 0-4

12 7-5 Applying the Systematic Treatment of Eq. 7.3 Question: What will e [Ca +, [F -, [HF if ph=3.00? [H + =.0 x 0-3 M - =.0x0 - M thus, from equation [HF. 5x0 [F How many [F - are converted to [HF? 60. % 7-5 Applying the Systematic Treatment of Eq Soluility of HgS When HgS dissolves in water, HgS (s) Hg + + S - sp =5 x 0-54 S - reacts with water since S - is a strong ase. S - + H O HS - + OH - = 0.80 HS - + H O H S + OH - =. x 0-7 water always follow H O H + + OH - w =.0 x 0-4

13 7-5 Applying the Systematic Treatment of Eq. 7.5 charge alance: [Hg + + [H + = [S - + [HS mass alance: [Hg + = [S - + [SH - + [H S In case of knowing ph of the solution or not, we always say that - = w /[H Applying the Systematic Treatment of Eq. 7.6 ) [H S? [HS [HS [S [HS [HS - in at eq 9-8 is originated from 9-7. Thus, [H S [S [H S [S ) [HS - directly [S [ HS

14 7.7 3) [S - directly Now, Eq 9-5 ecomes [S [S [S [S [Hg All of these equations must meet with Eq.9-6 in any case. [Hg [Hg [S [Hg sp 7-5 Applying the Systematic Treatment of Eq. 7.8 / sp [Hg / sp [S / sp [HS / sp S [H In case of ph 8, - =.0 x 0-6 M, sp,, are constants. All Calculations are dependent on - and fixed constants. 7-5 Applying the Systematic Treatment of Eq.

15 Ch. advanced Topics in Equilirium General approach to acid-ad systems Consider a solution : 0.0 mmol sodium tartarate (Na + HT - ) 5.0 mmol pyridinium chloride (PyH + Cl - ) 7.30 Charge alance mass alance [Na + = M [ + = M, [Cl - = M [H T + [HT - + [T - = M [PyH + + [Py = M Step : write a fractional composition eq. Step : sustitute fractional comp. expressions into charge alance. - = w /[H + Step 3: solve [H +

16 7.3 For monoprotic system: For diprotic system: 7.3 [H + + [PyH + + [Na + + [ + = - + [HT - + [T - + [Cl -

17 Dependence of Soluility on ph. Soluility of CaF 7.34 Charge alance: Mass alance: [total calcium species = [total fluoride species [Ca + = [HF= [CaOH + = [CaF + =

18 7.35 Homework 7.36 Ch7: 4, 8,, 8