Equilibrium

ABYSS Learning

E x . 1 6 Fluoroacetic acid has a K of 3.6 × 10–3. What concentration of the acid is needed so that [H+] is 2.0 × 10–3 ?
a

Sol. HC H FO + H O  C H FO – + H O+
22 2 2 22 2 3

[H O  ][C H FO  ] (2.0  10 3 )2
2
K = 3 2 2 = = 3.6 × 10–3

a [HC2H2FO2 ] [HC2H2FO2 ]

Thus [HC H FO ] = (2.0 103 )2 = 1.11 × 10–3 M remaining in solution.
22 2
(3.6 103 )

Total concentration

= (2.0 × 10–3) + (1.1 × 10–3) = 3.1 × 10–3 M

E x . 1 7 K of base imidazole at 25°C is 8.8 × 10–8.
b
(a) In what amounts should 0.02 M HCl and 0.02 M imidazole be mixed 100 mL of a buffer at pH = 7?

(b) When the resulting buffer is diluted to one litre, calculate pH of the diluted buffer.

Sol. (a) As pH = 7, pOH = 14 – 7 = 7 (at 25°C), pK = –log K = – log (8.8 × 10–8) = 7.0555
bb

Applying

[Salt]
pOH =pK + log

b [Base]

7 = 7.0555 + log [Salt]
[Base]

[Salt]
log = – 0.0555

[Base]

Taking antilog, [Base] = 1.14

[Salt]

or millimole of base = 1.14 .....(1)
millimole of salt

Suppose V ml of HCl is mixed with V ml of imidazole (base) to make the buffer.
12

millimole of HCl = 0.02 V1

millimole of imidazole = 0.02 V
2

As the buffer is of the base and its salt, 0.02 millimole of HCl will combine with 0.02 millimole of base

to give 0.02 millimole of salt.

 millimole of salt = millimole of HCl

= 0.02 V
1

and m.m. of base left = 0.02 V – 0.02 V
21

 From (1), we get, 0.02(V2  V1 ) = 1.14
0.02 V1

o r V2  V1 = 1.14 .....(2)
V1

Given that V + V = 100 mL .....(3)
12

From (2) and (3) we get, V1 = 31.84 mL and V2 = 68.15 mL

(b) pH shall remain the same on dilution as both K and [salt]/[base] will not change.
b

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E x . 1 8 A solution is prepared by mixing 200 mL of 0.025 CaCl and 400 mL of 0.15 M Na SO . Should CaSO
2 24 4

precipitate from a solution ?

Given K = 2.4 × 10–5
sp

S o l . The equation for the equilibrium is

CaSO (s)  Ca2+ (aq) + SO 2– (aq)
4 4

and the solubility product expression is

K = [Ca2+] [SO 2–] = 2.4 × 10–5
sp 4

If we assume that the volumes of the solutions that are mixed are additive, the final solution will have a volume

of 600 ml. This total volume contains the equivalent of 200 mL of CaCl , so the concentration of Ca2+ ions is
2

[Ca2+] =  200 mL CaCl 2 solution  × (0.025 M) = 8.33 × 10–3 M and the concentration of SO 2– ions is
 600 mL tota volume  4
l

[SO 2–] =  4 00 m L N a2SO4 solutio n  × (0.15 M) = 0.1 M
4  600 m L total volume 
 

The ionic product is

[Ca2+] [SO2–] = (8.33 × 10–3) (1.0 × 10–1) = 8.33 × 10–4
4

Which is larger than K , so CaSO should precipitate from the solution.
sp 4

E x . 1 9 Calculate the pH of a solution of 0.10 M acetic acid. Calculate the pH after 100 mL of this solution is treated

with 50.0 mL of 0.10 M NaOH. (K aCH3COOH  1.8 105 )

Sol. HC H O + HO  H O+ + C H O–
232 2 3 232

[H O  ] [ C H O  ]
2
K = 3 2 3 = 1.8 × 10–5

a [HC2H3O2 ]

Before treatment :

[H O+] = [C H O –] = X
3 232

[HC H O ] = 0.10 – X  0.10
232

x2 = 1.8 × 10–5 thus
0.10

x = 1.35 × 10–3 = [H3O+] and pH = 2.87

CH COOH + NaOH    CH COONa + HO
3 3 2
–
100 × 0.1 50 × 0.1 – –
–
= 10 =5 –

50 5

this is the buffer solution

[salt] [5]
pH = pk + log = 4.74 + log = 4.74
a [acid] [5]

E x . 2 0 Ionic product of water and ionization constant of acetic acid at 25°C are 1 × 10–14 and 1.75 × 10–5 respectively.
Calculate the hydrolysis constant of sodium acetate and its degree of hydrolysis in 10–3 M solution. Also calculate
the pH of the solution ?

S o l . CH3COONa is salt of weak acid and strong base ; its degree of hydrolysis may be calculated using the formula :

h=  Kh  =  Kw  ......(i)
 C   
 CKa 

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Hence, from Eq. (i)

h= 10 14 = 7.55 × 10–4
103  1.75  10 5

K (hydrolysis constant) = Kw 10 14 = 5.7 × 10–10
h Ka = 1.75 105

pH after salt hydrolysis may be calculated as -

1 .....(ii)
p H = [pK + pK + log C]
2w a

pK = – log K = – log 10–14 = 14
w w

pK = – log K = – log (1.75 × 10–5) = 4.7569
a a

log C = log 10–3 = – 3

Substituting the values in Eq.(ii), we get

1
p H = [14 + 4.7569 – 3]

2

or pH = 7.88

E x . 2 1 If [Cd2+] = [Zn2+] = 0.1 M initially, what [H+] must be maintained in a saturated H S (0.1 M) to precipitate CdS
2
but not ZnS ?

K (CdS) = 8 × 10–27
sp

K (ZnS) = 1 × 10–21
sp

K (H S) = 1.1 × 10–21
a2

Sol. In order to prevent precipitation of ZnS,

[Zn2+] [S2–] < K (ZnS) = 1 × 10–21
sp

(Ionic product)

or (0.1).[S2–] < 1 × 10–21

o r [S2–] < 1 × 10–20

This is the maximum value of [S2–] before ZnS will precipitate. Let [H+] to maintain this [S2–] be x.

Thus for HS  2H+ + S2–,
2

K = [H  ]2 [S2– ] = x 2 (1  10 –20 ) = 1.1 × 10–21
a [H2S] 0.1

or x = [H+] = 0.1 M.

 No ZnS will precipitate at a concentration of H+ greater than 0.1 M.

E x . 2 2 If 0.10 M KH BO is titrated with 0.10 M HCl, what indicator should be used for this titration? [K 7.3 × 10–10]
23 a

Sol. H BO – + H O+  H BO + HO
23 3 33 2

At the equivalence point, 0.050 M H3BO3 would be produced. Only the first ionization step of H3BO3 is

important to the pH.

H BO + HO  H BO – + H O+
33 2 23 3

[H O  ][H B O  ] x2
3
K = 3 2 =

a [H3BO3 ] 0.050

= 7.3 × 10–10 thus x = 6.0 × 10–6 and pH = 5.22
pH 5.22 is in the middle of the range of methyl red, which would therefore be suitable.

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E x . 2 3 In an attempted determination of the solubility product constant of T S, the solubility of this compound in pure
2
CO free water was determined as 6.3 × 10–6 mol/L. Assume that the dissolved sulphide hydrolysis almost
2
completely to HS– and that the further hydrolysis to H S can be neglected. What is the computed K ?
2 sp
(K = 1.0 × 10–14)
2

Sol. T S  2T + + S2– , K = [T +]2 [S2–]
2 sp

S2– + HO  HS– + OH–
2

K = Kw 1.0 1014 = 1.0
h K2 = 1.0 1014

[T +] = 2(6.3 × 10–6) M, [S–2] = 6.3 × 10–6 M, and since [HS–] = [S–2]

K= (6.3 10 6 )2 = 1.0
h
[S 2 ]

[S2–] = (6.3 × 10–6)2

K = (6.3 × 10–6)2 [2(6.3 × 10–6)]2
sp
= 6.3 × 10–21

E x . 2 4 To reduce [Cu2+] to 10–12 how much NH should be added to a solution of 0.0010 M Cu(NO ) ? Neglect the
3 32
amount of copper in complexes containing fewer than 4 ammonia molecules per copper atom.

(K Cu(NH ) 2+= 1 × 10–12)
d 34

Sol. Cu(NH ) 2+  Cu2+ + 4NH
34 3

K = [Cu2 ][N H 3 ]4 = 1.0 × 10–12
d [Cu(NH 3 )24 ]

since the sum of the concentration of copper in the complex and in the free ionic state must equal 0.0010 mol/L,
and since the amount of the free ion is very small, the concentration of the complex is taken to be 0.0010 mol/L.

Let x4 = [NH ]
3

Then (10 12 ) (x4 ) = 1.0 × 10–3
0.0010

or x4 = 1.0 × 10–2

or x = 0.178

The concentration of NH at equilibrium is 0.178 mol/L. The amount of NH used up in forming
33

0.0010 mol/L of complex is 0.0040 mol/L, an amount negligible compared with the amount remaining at
equilibrium. Hence the amount of NH to be added is 0.178 mol/L.

3

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Ex 1. SOLVED PROBLEMS (SUBJECTIVE)
Sol.
The solubility of Pb(OH) in water is 7.6 × 10–6 M. Calculate the solubility of Pb(OH) in buffer solution of pH
Ex 2. 22
Sol.
= 8.

K =spPb(OH )2 4S3 = 4 × (7.6 × 10–6)3 = 1.755 × 10–15
The pH of buffer solution = 8, pOH = 6
 [OH–] = 10–6

For left solubility of Pb(OH)
2

[Pb2+] [OH–]2 = K
sp

 (S)(2S + 10–6)2 = 1.755 × 10–15

1.755  10 15 (2S + 10–6  10–6)
 S = 10 12

 S = 1.755 × 10–3 mol/L

Calculate pH of the following mixtures. Given that K = 1.8 × 10–5 and K = 1.8 × 10–5:
ab

(a) 50 mL of 0.05 M NaOH + 50 mL of 0.10 M CH COOH
3

(b) 50 mL of 0.10 M NaOH + 50 mL of 0.10 M CH COOH
3

(a) CH COOH + NaOH   CH COONa + HO
3 3 2

Initial 50 × 0.1 50 × 0.05 0 0

Millimoles =5 2.5

Millimoles

after reaction 2.5 0 2.5 2.5

The solution consists of CH COOH and CH COONa and thus for buffer
33

[Salt]
pH = pK + log

a [Acid]

pH = pKa + log [C H 3C O O Na ]
[C H 3C O O H ]

2.5 / 100
 pH =– log 1.8 × 10–5 + log 2.5 / 100    pH = 4.7447

(b) CH COOH + NaOH   CH COONa + HO
3 3 2

Initial millimoles 50 × 0.1 50 × 0.1 0 0

=5 =5 0 0

Final millimoles 0 05 5

5
Finally concentration of CH COONa = , and pH is decided by salt hydrolysis.

3 100

CH COONa + H O  CH COOH + NaOH
32 3

C 00

C(1 – h) Ch Ch

[OH–] = Ch = C Kh = Kw C = 1 0 14 5 = 5.27 × 10–6 M
C Ka 1.8 105 100

[H+] 1 0 14 1 0 8 = 0.189 × 10–8 or pH = 8.72
= 5.27 106 =

5.27

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Ex 3. Calculate the pH of an aqueous solution of 0.2 M ammonium formate assuming complete dissociation (pK
Sol. a
Ex 4.
Sol. of formic acid = 3.8 and pK of ammonia = 4.8)
b
Ex 5.
Sol. HCOONH + H O  HCOOH + NH OH 4
42
Ex 6.
Sol. The pH of the salt of weak acid and weak base is given by

11 = 6.5
pH = [pK + pK – pK ]  pH = 2 [14 + 3.8 – 4.8]
2w a b

Should Mg(OH) precipitate from a solution that is 0.001 M MgCl if the solution is also made 0.10 M in
22

NH3 [K sp[Mg(OH )2 ]  1.8  10 11 , K b(NH4OH )  1.8  10 5 ] .

[OH–] in 0.10 M NH OH = K bC (Ostwald's dilution law)
4

 [Mg2+] = 1.8 105  0.1 = 1.34 × 10–3 M
= 0.001 M

Ionic product = [Mg2+] [OH–]2 = (0.001) × (1.34 × 10–3)2

= 1.8 × 10–9 > K
sp

As, Ionic product is greater than K of Mg (OH) , hence precipitation should occur.
sp 2

Calculate the pH at which an acid indicator with K = 1 × 10–5 changes colour when indicator concentration
a

is 1 × 10–3 M. Also report the pH at which coloured ions are 60% present.

For indicator dissociation equilibrium

Hl  H+ + l–
n n

Colour A Colour B

[H  ] [l  ]
n
K =
ln [Hln ]

The mid - point of the colour change of an indicator Hl is the point at which
n

[ln–] = [Hl ], K = [H+] = 1 × 10–5
n ln

 pH = 5

Thus, at pH = 5 the indicator will change its colour.

[H  ] [l  ] [H] 60 / 100
n 20 /100
K =  1 × 10–5 =
ln [Hln ]

 [H+] = 0.666 × 10–5

 pH = 5.1760

A solution has 0.1 M Mg2+ and 0.05 M NH . Calculate the concentration of NH Cl required to prevent the
34

formation of Mg(OH) in sol ut io n. K sp =18 .0 × 10–12 and ionisation constant of NH is 1.8 × 10–5.
2 3
M g ( OH )2 

The minimum [OH–] at which there will be no precipitation of Mg(OH) can be obtained by K = [Mg2+] [OH–]2
2 sp

 18.0 × 10–12 = (0.1) × [OH–]2

 [OH–] = 1.34 × 10–5 M

Thus, solution having [OH–] = 1.34 × 10–5 M will not show precipitation of Mg(OH) in 0.1 M Mg2+. These
2

hydroxyl ions are to be derived by basic buffer of NH Cl and NH OH.
44

pH = pK + log [Salt]  pH = pK + log [N H  ]
b [Base] b 4

[NH 4 O H ]

NH OH  NH + + OH–
4 4

In presence of [NH Cl], all the NH + ions provided by NH Cl as due to common ion effect, dissociation of
44 4

NH OH will be suppressed.
4

–log [OH–] = – log 1.8 × 10–5 + log [N H  ]
4

[0.05]

 [NH +] = 0.067 M or [NH Cl] = 0.067 M
44

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Ex 7. What is pH of 1 M CH COOH solution? To what volume must one litre of this solution be diluted so that the
Sol. 3

Ex 8. pH of resulting solution will be twice the original value. Given : K = 1.8 × 10–5
Sol. a

H CCOOH + H O  H CCOO– + H O+
3 2 33

Initial 1M 00

–xM xM xM

Final (1–x) M xx

K= x2 x2  x = = 4.2 × 10–3 = [H O+]
a  Ka 3
1x 1

pH = – log [H O+] = – log {4.2 × 10–3} = 3 – log 4.2 = 2.37
3

Now, let 1 L of 1 M AcOH solution be diluted to V L to double the pH and the conc. of diluted solution be
C.

New pH = 2 × Old pH = 2 × 2.37 = 4.74

pH = – log [H O+] = 4.74
3

 [H O+] = 1.8 × 10–5
3

H CCOOH + HO  H CCOO– + H O+
3 2 3 3
0 0
Initial C 1.8 × 10–5
1.8 × 10–5
–1.8 × 10–5

Final C–1.8×10–5 1.8 × 10–5 1.8 × 10–5

K= [CH3COO– ]  [H3O ]
a [C H 3 C O O H ]

1.8 × 10–5 = 1.8  10 5  1.8  10 5
C  1.8  10 5

 C = 3.6 × 10–5 M

On dilution,

MV = MV
11 22

1 M × 1 L = 3.6 × 10–5 M × V
2

 V = 2.78 × 104 L
2

A sample of AgCl was treated with 10 mL of 1.7 M Na CO solution to give Ag CO . The remaining
23 23

solution contained 0.0026 g of Cl– per litre. Calculate solubility product of AgCl. K sp(Ag2CO3 ) = 8.2 × 10–12

[CO32–] = [Na2CO3] = 1.7 M

At eqm., [Cl–] = [NaCl] 0.0026 = 7.32 × 10–5 M
=
35.5

2AgCl (s) + Na CO  Ag CO (s) + 2NaCl
23 23

Initial 1.7 M 0

At eqm. (1.7 – 7.32 × 10–5) 7.32 × 10–5 M

[Ag+]2 [CO ] = K2–
3 sp( Ag2CO3 )

 [Ag+] = K sp( Ag2CO3 ) = 8.2 10 12 = 2.1963 × 10–6 M
1.7
C O 2 
3

 K =sp(AgCl) [Ag+] [Cl–] = (2.1963 × 10–6) × (7.32 × 10–5) = 1.61 × 10–10

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Ex 9. Given Ag+ (NH )  Ag+ + 2NH , K = 8.2 × 10–8 and K =sp(AgCl) 2.378 × 10–10 at 298 K. Calculate the
Sol. 32 3c

concentration of the complex in 1.0 M aqueous ammonia.

[Ag(NH ) ] +(aq)  Ag+ (aq) + 2NH (aq)
32 3

x+y 2x

AgCl (s)  Ag+ (aq) + Cl– (aq)

x+y y

In case of simultaneous solubility, Ag+ remains same in both the equilibrium

K= (x  y)  (2x)2 .......(1)
c
[Ag(NH3 )2 ]

K = (x + y) × y .......(2)
sp

 Kc (2 x )2 Given, [NH ] = 2x = 1 M
K sp = [Ag(NH3 )2 ]  y 3

[Ag(NH ) ]+ = [Cl–] = y because Ag+ obtained from AgCl passes in [Ag(NH ) ]+ state.
3 2 32

Kc = 1  y2 = 2.378 10 –10 = 0.29 × 10–2  y = 0.539 × 10–1 = 0.0539 M
K sp y  y 8.2 108

That is, [Ag(NH ) ]+ = 0.539 M
32

Ex 10. How many moles of NH must be added to 1 litre of 0.1 M AgNO solution to reduce Ag+ concentration
Sol. 33

to 2 × 10–7 M. K [Ag(NH ) ]+ = 7.2 × 10–8
d 32

As K [Ag(NH ) ]+ = 1 = very-very large
f 3 2 7.2  10 8

Hence, almost all Ag+ ions will be converted to [Ag(NH ) ]+
32

 [Ag(NH ) ]+  0.1 M
32

[Ag+] = 2 × 10–7

K = [Ag  ][NH 3 ]2  7.2 × 10–8 = 2  107  [NH 3 ]2
[Ag(NH3 )2 ] 0.1

[NH ] = 0.189 M
3

It is the concentration of free NH .
3

[NH3]total = [NH3]free + [NH ]3 complexed = 0.189 + 2 × 0.1 = 0.389 M

Ex 11. (i) What mass of Pb2+ ion is left in solution when 50 mL of 0.2 M Pb(NO ) is added to 50.0 mL of
Sol. 32
1.5 M NaCl ? (K PbCl = 1.7 × 10–4)
sp 2

(ii) 0.16 g of N2H4 is dissolved in water and the total volume made up to 500 mL. Calculate the

percentage of N H that has reacted with water at this dilution. The K for N H is 9.0 × 10–6 M.
24 b 24

(i) Millimoles of Pb2+ before precipitation = 50 × 0.2 = 10

Millimoles of Cl– before precipitation = 50 × 1.5 = 75

Assuming complete precipitation of PbCl followed by establishment of equilibrium.
2

Millimoles of Cl– left after precipitation

= 75 – 20 = 55 in 100 mL.

After precipitation [Cl–] = 0.55 M

That means, we have to find out solubility of PbCl in 0.55 M Cl– ion solution.
2

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PbCl  Pb2+ + 2Cl–, K = [Pb2+] [Cl–]2
2 sp

[Pb2+] = K sp = 1.7 10 4 = 5.6 × 10–4 M
[Cl ]2 (0 .5 5 )2

Mass of Pb2+ in solution = 5.6  10 4
× 100 × 208 = 1165 mg
1000

(ii) [N H ] = 0.16 × 1000 0.01 M
=
2 4 32 500

NH + HO  N H+ + OH–
24 2 25

1 00

(1–) 

K = C2,
b

2 = K b 9.0  10 6 = 9 × 10–4
=
C 0.01

  = 9.0 104 = 3 × 10–2 = 3 %

Ex 12. If very small amount of phenolphthalein is added to 0.15 M solution of sodium benzoate, what fraction of
Sol.
the indicator will exist in the coloured form ?

K = 6.2 × 10–5
a(Benzoic acid)

K = 1 × 10–14
w(H2O)

K = 3.16 × 10–10
ln(Phenolphthalein)

Formula for pH of solution due to hydrolysis of C H COONa
65

1 + pK a + log C] 1 [14 – log 6.2 × 10–5 + log 0.15] = 8.6918
pH = [pK =
2w 2

Formula for pH of indicator

pH = pK + [I n]  8.6918 = – log (3.16 × 10–10) + [I n]
In log log [H In ]

[H In ]

[I n]
 = 0.16 (Fraction of indicator in coloured form = 0.16)

[H In ]

Ex 13. What will be the Ag+ ion concentration in a solution of 0.2 M solution of [Ag(NH ) ]+ ?
Sol. 32

Ex 14. Ag(NH ) +  Ag+ + 2NH ; K = 5.8 × 10–8
32 3 dis

Let, concentration of Ag+ at equilibrium be C

 [NH ] = 2C
3

K= [Ag  ][NH 3 ]2  5.8 × 10–8 = C  4C2
dis [Ag(NH 3 )2 ] 0.2

 C = 0.0014 M.

A solution contains 0.1 M Cl– and 0.001 M CrO 2–. If solid AgNO is gradually added to this solution which
43

will precipitate first, AgCl or Ag CrO ? Assume that the addition causes no change in volume. Given
24

K = 1.6 × 10–10 M2 and K = 1.79 × 10–12 M3. What % of Cl– remains in solution when CrO2–
sp(AgCl) sp(Ag2CrO4) 4

starts precipitating ?

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Sol. Ag+ ion concentration required for precipitation

For AgCl, [Ag+] = K sp(AgCl) 1.6  1010 = 1.6 × 10–10 M
[Cl  ] =
0.1

For Ag CrO , [Ag+]2 = K sp( Ag2CrO4 ) 1.79 10 12
24 =
[ C rO 2 ]
4 (0.001)

 [Ag+] = [1.79 × 10–9 M2]½ = 4.23 × 10–5 M

AgCl will precipitate first because it requires low concentration of Ag+. Remaining concentration of Cl–

when Ag CrO starts precipitating = K sp(AgCl) 1.6 1010 = 3.78 × 10–6 M.
24 [Ag ] = 4.23 105

% of remaining concentration of Cl– 3.78 106 × 100 = 3.78 × 10–3 %
=
0.1

Ex 15. Determine the concentration of NH solution whose one litre can dissolve 0.10 mol of AgCl. K and
Sol. 3 sp(AgCl)

K fAg(NH3 )2  are 10–9 M2 and 1.6 × 106 M–2 respectively.

AgCl (s)  Ag+ + Cl– K =K
1 sp
Ag+ + 2NH  Ag+(NH )
3 32 K =K
2f

AgCl (s) + 2NH  Ag+ (NH ) + Cl– K=K ×K
3 32 sp f

[ A g(NH3 )2 ] [C l  ]
[NH 3 ]2
 K = (Given solubility of AgCl = 0.10)

 [Ag(NH ) ]+ = 0.10 M,
32

Also, [Cl–] = 0.1

1 × 10–9 × 1.6 × 106 = 0.1  0.1
[NH 3 ]2

 [NH ]2 = 6.25  [NH ] = 2.5 M
3 3

Thus, [NH ] at equilibrium = 2.5 M
3

Also 0.2 M of NH must have been used to dissolve 0.1 M AgCl
3

 [NH ] = 2.5 + 0.2 = 2.7 M
3 Total

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