Summary of Ionic Equilibrium

8/2/2019 Summary of Ionic Equilibrium

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General procedure for solving pH/pOH problems involving strong acids/bases: write down full dissociation equation for strong acid/base

o strong monoacid: HX(aq) ~ H'(aq) + X'(aq)o strong monobase:YoH(aq) ~ Y+(aq) + OH-(aq}

determine H+/OH- concentration using mole ratio consider water dissociation equilibrium for dilute solution with concentration < 10-5mol dm-3 determine pH/pOH using formula

o pH formula: pH = -lg[Hj = > [H+] = 10-pHo pOH formula: pOH = -lg[OHl = > [OHl = 10-pOHo relationship at 25C: pH + pOH :::;14 from K w = [H+][OHl = 10-14 mol2 dm-6

Ex (a) Calculate the pH of 0.2 mol dm -3H2S04H2S04(aq) ~ 2H+(aq) + $042-(aq)H2$04",,2H+[H+j = 2,x0.2 = 0.4 mol drn?pH = -lg[Hj = -lg(O.4) = 0.40

(b) Calculate the pH of 4 x 10-8mol dm-3 Ca(OH)2Ca(OHk(aq) ~ Ca2+(aq) + 20H-(aq)Ca(OHh .. 20H-[Ca(OH)21 < 10-Smol dm-3is very low = > consider H20dissoc. eqm.[OHl = [OH1Ca(oH)2+ [OH1H20:::;2 x 4 X 10-8+ 10-7:::; 1.8 x 19-7 mol dm-3pOH =-lg[OHl =-lg(1.8 x 10-7) = 6.74pH:::; ,14- pOH = 14 - 6.74 = 7.26

2 General procedure for solving pH/pOH problems involving weak acids/bases: write down partial dissociation equati~n for weak acid/base

o weak monoacid: HA(aq) :> H'(aq) + A'(aq)o weak monobase: 8(aq) +HzO(I) ee - BH+(aq) + OH-(aq)

write down KJKb expression for weak acid/base

o CO [H+J[]monoacid diSSOCiationconstant: K.= ,o ' 0 [HA]

o [BH+][OH-]monobase dissociation 'constant: ~ = [B ]


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1 General procedure for solvrtiQ pH/pOH problems involving strong acids/bases:wr ite down ful l dissociation equat ion for strong acid/baseo strong monoacid: HX(aq) ~ H'(aq) + X(aq)o strong monobase:YOH(aq) ~ Y'(aq) + OH-(aq)

determine H+{OH- concentration using mole ratio ~.

consider water dissociation equilibrium for dilute solution with concentration < 10-5mol dm-3 determine pH/pOH using formula

o pH formula: pH = -lg[Hj = > [H+j = 10pHo pOH formula: pOH = -lg[OHl : : : : > [OHl = 10pOHo relationship at 25C: pH + pOH = 14 from K w = [H+UOHl = 10-14 mol2 dm"

Ex (a) Calculate the pH of 0.2 mol dm-3H2S04H~04(aq) ~ 2H"(aq) + 8042-(aq)H2S04 .. 2H+[Hj = 2:x 0.2 = 0,4 mol dm-3pH = -lg[Hj = -lg(O.4) = 0.40

(b) Calculate the pH of 4 x 10"" mol dm-3 Ca(OH)2Ca(OHMaq) ~ Ca2+(aq) + 20H-(aq)Ca(OHh '" 20H'[Ca(OH)i1 < 10-5mol dm-3is very low = > consider H20 dissoc. eqm.[OHl = [OH1Ca(OH)2 [OH1H2o=2x 4 x10-a +10-7 = 1.8 x 10-7 mol dm-3pOH = -lg[OHl =-lg(1.8 x 10-7) = 6.74pH = 14 - pOH = 14 - 6.74 = 7.26

2 General procedure for solving pH/pOH problems involving weak acids/bases: write down partial dissociation equatic:m for weak acid/base

o weak monoacid: HA(aq) ~ H+(aq) + A-(aq)o weak monobase: B(aq) +H20(1)~ BH+(aq) + OH-(aq)

write down KJKb expression for weakacid/base

o .' . [H+J[]monoacid dissociation constant: K a . = .. . . [1M]

o [BH+][OH-]monobase dissociation constant: K t. = .[B ]


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o relationship at 25C for conjugate acid-base pair: KaKb = Kw = 10-14 rnor'drn" define equilibrium concentrations involved using equilibrium table substitute equilibrium concentrations into Ka/Kb expression and solve using either one of the

two approachesKa Ko approximate approach when or b s2.5 x 10-3: [HA] ~[HA]inilial or [B] ~ [B];nilial

[H Alllitial [B L'itialo exact approach when K; or K; > 2.5 X 10-3[H Allli lial [B Litial

determine pH/pOH using formula as for strong acids/basesEx (c) Calculate the pH of 0.1 mol dm" C2HsC02H with Ka= 1.45 X 10-s mol dm?

Let eqm [H+] be x mol dm"C2H5C02H(aq) < = > H'(aq) + C2H5C02-(aq)

Initial [ ]: 0.1 0 0fI . in [ ]: -x +x +xEqm. []: 0.1 - x x xAssume x is so small that eqm. [C2H5C02H] "" 0.1 mol dm"Ka = [H+] [C~HsCO~ - ]

[C1HsC01H]7x- 5= 1.45 x 10-0.1

x = 1.20 X 10-3pH = -lg[H+] = -lg(1.20 x 10-3) = 2.91

(d) Calculate the pH of 0.5 mol drn" CH3NH2 with Kb = 4.5 X 10-4mol dm-3Let eqm [OHl be x mol drn"

Initial [ ]: 0.5 o ofI. in []: -x +x +xEqm. []: 0.5-x x xAssume x is so small that eqm. [CH3NH2] ~ 0.5 mol drn"

x = 0.015pOH = -lg[OHl = -lg(0.015) = 1.82pH = 14 - pOH = 14 - 1.82 = 12.2

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(e) Calculate the pH of 0.3 mol dm" NaOCI given Ka of HOCI = 4 x 10-8 mol dm"K . 10-14

K b of conjugate base OCI" = _" = 8 = 2.5 x 10.7 mol dm3, 4xlO-Let eqm [OHl be x mol drn"

OCI-(aq) + H20(I) < = > HOCI(aq) + Ol+jaq)Initial [ ]: 0.3 o ob. in [ ]: -x +x +xEqm. []: 0.3 - x x xAssume x is so small that eqm. [OCll "" 0.3 mol dm"K - [HOCI][OH-] = x2 = 2.5 X 10-7b - [OCr] 0.3

x=2.74x10-4pOH = -lg[OHl = -lg(2.74 x 10 -4) = 3.56pH = 14-pOH = 14-3.56 = 10.4

(f) Calculate the pH of 0.6 mol dm? NH4CI given Kb of NH3= 1.6 X 10.5 mol dm"K 10-14Ka of conjugate acid NH/ = _H = = 6.25 X 10 -10 mol drn", 1.6 x 10-5

Let eqm [H+] be x mol drn"NH/(aq) < = > NH3(aq) + H'(aq)

Initial []: 0.6 0 0b. in []: -x +x +xEqm. []: 0.6-x x xAssume x is so small that eqm. [NH/] "" 0.6 mol drn?

x = 1.94xlO5pH = -lg[H+] = -Ig( 1.94x 10-5) = 4.71

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3 General remarks involving acid-base titrations: general titration curves involving acids/bases of 0.1 mol drn" concentrations

r e f i r~ 2 F ) l I r S C h M J Sf { c ' f u t ( . e _ (ly feS

determine appropriate indicator with pH transition range within end-point rapid pH changetype of titration end-pt rapid pH change indicator pH range

strong acid-strong base ::::3-11 any below see belowstrong acid-weak base ::::3-7 methyl orange ~ 3.2-4.4weak acid-strong base "" 7-11 phenolphthalein "" 8-9.8

4 General procedure for solving pH/pOH problems involving buffer solutions: determine nature of acidic/basic buffer solution

o acidic buffer pair: weak acid HA and salt MA of conjugate base (A)o basic buffer pair: weak base B and salt BHX of conjugate acid (BH+)write down pH/pOH expression for acidic/basic buffer solution

o pH expression for acidic buffer pair: pH = pK, + 19 [A-] ~. [RAJ C a v ' d J

c : S t f 1 1 JCha.(0

o [BR+]pOH expression for basic buffer pair: pOH = pKb + Ig - = - _ _[B ] define equilibrium concentrations involved using given concentrations of buffer pair

o acidic buffer pair: [A1""MA] and [HA] "" [HA]initialo basic buffer pair: [BH+] ~ [BHX] and [B] "" [B]inilial

substitute equilibrium concentrations into pH/pOH expression and solve

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Ex (g) Calculate the pH of buffer solution made up of aqueous 0.7 mol dm? CH3C02H (with Ka =1.8 X 10-5mol dm") and 0.5 rnoldm" C2H5C02Na

(h) Calculate the pH of buffer solution prepared by mixing equal volumes of aqueous 0.6 moldrn" NH3(with Kb = 1.6 X 10-5mol dm") and 0.8 mol drn" NH4N03Upon mixing equal volumes: [NH3]= 0.6/2 = 0.3 mol drn" and [NH4N031 = 0.8 / 2 = 0.4 mol dm?

[NH/] -5 0.4pOH=pKb+lg =-lg(1.6x10 )+Ig- =4.92[NHJ 0.3pH = 14 - pOH = 14 - 4.92 = 9.08

5 General procedure for solving pH/pOH problems involving buffer solutions upon adding smallamounts of acids/bases:

determine H+/OH-concentration for acid/base addedto buffer solution write down neutralisation equation between added acid and weak/conjugate base or added

base and weak/conjugate acid to maintain fairly constant pH/pOHo addition of acid to acidic buffer pair: H'(aq) + A(aq) -+ HA(aq)

H'(aq) + B(aq) -+ BH+(aq)addition of acid to basic buffer pair:o addition of base to acidic buffer pair: OH-(aq) + HA(aq) -+ A'(aq) + H20(I)

OH-(aq) + BH+(aq) -+ B(aq) + H20(I)addition of base to basic buffer pair: redefine new equilibrium concentrations based on above neutralisation equation substitute new equilibrium concentrations into pH/pOH expression and solve

Ex (i) Calculate the new pH of 1 dm" of the buffer solution in (g) in each case when(1) 0.05 mol of NaOH is added

[OHl added = 0.05 mol drn"

Adj. eqm. []:~ in []:

0.7 0.05 0.5-0.05 -0.05 +0.05

New eqm. []: 0.65 o 0.55[CH3CO?-] -5 0.55New pH= pKa+ Ig - = -lg(1.8 x 10 ) + Ig-- = 4.67[CH3C02H] 0.65

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(2) 0.01 mol of H2S04 is added[H+]added = 2 x 0.01 = 0.02 mol drn"On adding H+: CH3C02-(aq) + H'(aq) -+ CH3C02H(aq)Adj. eqm. []: 0.5 0.02 0.7t. in []: -0.02 -0.02 +0.02New eqm. []: 0.48 o 0.72

[CH3CO? -] -5 0.48New pH = pKa+ Ig - = -lg(1.8 x 10 ) + Ig-- = 4.57[CH3C02H] 0.72(j) Calculate the new pH of 1 dm" of the buffer solution in (h) in each case when

(1) 0.02 mol of Ca(OHh is added[OHl added = 2 x 0.02 = 0.04mol drn"On adding OH-: NH/(aq) + Ol-l'(aq) -+ NH3(aq) + H20(I)Adj. eqm. []: 0.4 0.04 0.3t. in [ ]: . -0.04 -0.04 +0.04New eqm. []: 0.36 o 0.34

[NH+] 0.36New pOH = pKb+ Ig 'I = -lg(1.6 x 10-5) + Ig-- = 4.82[NH3] 0.34New pH = 14- pOH = 14 - 4.82 = 9.18

(2) 0.03 mol of HCI is added[H+]added = 0.03 mol dm"On adding H+:NH3(aq) + H'(aq) -+ NH/(aq)Adj. eqm. []: 0.3 0.03 0.4t. in []: -0.03 -0.03 +0.03

New eqm. []: 0.27 o 0.43[NH. " J 0.43New pOH = pKb+ Ig 4 = -lg(1.6 x 10-5) + Ig-- = 5.00[NH3] 0.27

New pH = 14- pOH = 14 - 5.00 = 9.006 General procedure for solving solubility problems involving sparingly soluble salts:

write down partial dissociation equation for sparingly soluble salto sparingly soluble binary salt: MX(s) < = > M+(aq) + X'(aq)

write down Ksp expressiono solubility product for binary salt: Ksp= [M+][Xl

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define equilibrium concentrations involved based on solubility, i.e. maximum no. of mol. of saltdissolved per drn" of water substitute equilibrium concentrations into Kspexpression and solve

Ex (k) Calculate the pH of saturated aqueous Ca(OHh with Ksp = 1 X 10-11 mof dm"Let solubility of Ca(OH)z be x mol drn",

Initial []: x o o~ in [ ] at satn: -x +x +2xEqm. [] at satn: o x 2xKsp= [Ca2+][OH12 = X(2X)2 = 1 X 10-11 => X = 1.36 x 10-4[OHl = 2x = 2 x 1.36 X 10-4 = 2.71 X 10-4 mol dm"pOH = -lg[OHl = -lg(2.71 x 10-4) = 3.57pH = 14 -pOH = 14-3.57 = 10.4

(I) Calculate the Kspof Pbl2 (Mr = 461) with solubility of 0.46 g drn" at 15C0.46 4 3Solubility of Pbl2 = - - = 9.98 x 10- mol ern461PbI2(s) < = > Pb2+(aq) + 21-(aq)

Initial []: 9.98x10-4 0 0~ in [ ] at satn: -9.98x10-4 +9.98x10-4 +2(9.98x 10-4)Eqm. [] at satn: 0 9.98x10-4 2.00x10-3

7 General procedure for solving precipitation problems involving mixing reagents to form sparinglysoluble salts: determine sparingly soluble salt-to be formed based on the ions mixed write down ionic product expression (expression same as Kspbut not necessarily magnitude) define aqueous ion concentrations involved using given concentrations of reagents mixed substitute aqueous ion concentrations into ionic product expression and compare itsmagnitude with Ksp(a constant at constant temperature)

o ionic product < Ksp:below saturation point => no precipitation occurso ionic product = Ksp:at saturation point => onset of precipitationo ionic product> Ksp:beyond saturation point => precipitation occurs

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Ex (m) Predict if a precipitate of Mg(OHh (Ksp = 4 x 10-6 mol3 dm") would be formed in each casewhen(1) equal volumes of 0_01mol dm" MgCI2and 0.02 mol drn" Ca(OHh are mixed

To obtain Mg(OH)z on mixing aq MgCI2 and Ca(OHh1 1Ionic pdt = [Mg2+][OH12 = ( - x 0.01)( - x 2 x 0.02)2 = 2 x 10-6 mor' drn" < Ksp2 2

Below saturation point => ppt of Mg(OHh will not be formed(2) 80 ern" of 0.2 mol dm" Mg(N03h and 20 ern" of 0.3 mol dm" NaOH are mixed

To obtain Mg(OHh on mixing aq Mg(N03h and NaOH+ 80 20Ionic pdt = [Mg2 ][OH12 = (-x 0.2)(-x 0.3)2 = 5.76 x 10-4 mol" drn" > Ksp100 100

Beyond saturation point => ppt of Mg(OHh will be formed

Summary of Ionic Equilibrium

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