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Solubility Product
We have come across many precipitation reactions in inorganic qualitative analysis. For
example, dil HCl is used to precipitate Pb2+ ions as PbCl2 which is sparingly soluble in water. Kidney stones are developed over a period of time due to the precipitation of Ca2+ (as calcium oxalate etc…). To understand the precipitation, let us consider the solubility equilibria that exist between the undissociated sparingly soluble salt and its constituent ions in solution.
For a general salt X Y m n
, X Y (s) mX (aq) + nY (aq)
m n
H O n+ m-2 →←
The equilibrium constant for the above is
K = [X ] [Y ] [X Y ]
n+ m m- n
m m
In solubility equilibria, the equilibrium constant is referred as solubility product constant (or) Solubility product.
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In such heterogeneous equilibria, the concentration of the solid is a constant and is omitted in the above expression
K = [X ] [Y ]sp n+ m m- n
The solubility product of a compound is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co – efficient in a balanced equilibrium equation.
Solubility product finds useful to decide whether an ionic compound gets precipitated when solution that contains the constituent ions are mixed.
When the product of molar concentration of the constituent ions i.e., ionic product, exceeds the solubility product then the compound gets precipitated.
The expression for the solubility product and the ionic product appears to be the same but in the solubility product expression, the molar concentration represents the equilibrium concentration and in ionic product, the initial concentration (or) concentration at a given time ‘t’ is used.
In general we can summarise as, Ionic product > Ksp , precipitation will occur and the solution is super saturated. Ionic product < Ksp , no precipitation and the solution is unsaturated. Ionic product = Ksp , equilibrium exist and the solution is saturated.
Example 8.9
Indicate find out whether lead chloride gets precipitated or not when 1 mL of 0.1M lead nitrate and 0.5 mL of 0.2 M NaCl solution are mixed? Ksp of PbCl2 is 1.2 10
-5× .
PbCl Pb (aq)+2Cl (aq) Ionic product = [Pb ]
2 (s) H O 2+ -
2+
2 ⇀ ↽ [Cl ]- 2
Total volume = 1.5 mL
Pb NO Pb +2NO3 M
2+
M 3( ) − 2
0 1 0 1
. .
⇀ ↽
No of moles of Pb2+ = Molarity × volume of the solution in litre = 0.1×1×10–3 = 10–4
[Pb number of moles of Pb
Volume of the solution in L
2+
2+
]= = mL
= 6.7 10 M -210
1 5 10
4
3
−
−× ×
.
NaCl Na Cl M
M
-
M0 2 0 2 0 2. . . → +
No of moles of Cl– = 0.2 × 0.5 × 10–3= 10–4
[Cl moles 10 L
= 6.7 10 M-3 -2-]
. =
× ×
−10 1 5
4
Ionic product = (6.7 10 = 3.01 10-2 -4× × ×−)( . )6 7 10 2 2
Since, the ionic product 3.01 10 4× - is greater than the solubility product (1.2 10-5× ) ,
PbCl2 will get precipitated.
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Determination of solubility product from molar solubility
Solubility product can be calculated from the molar solubility i.e., the maximum number of moles of solute that can be dissolved in one litre of the solution.
For a solute X Y ,m n
X Y (s) mX (aq) +nY (aq)m n n+ m-
From the above stoichiometrically balanced equation we have come to know that 1 mole of X Y (s)m n dissociated to furnish ‘m’ moles of Xn+ and ‘n’ moles of Ym- if ‘s’ is molar solubility of X Y ,m n then
[X ]=ms and [ Y ]=ns K = [X ] [Y ]
K =(
n+ m-
sp n+ m m- n
sp
∴ ms) (ns)
K =(m) (n) (s)
m n
sp m n m+n
Example 8.10
• Establish a relationship between the solubility product and molar solubility for the following
a) BaSO b) Ag (CrO )4 2 4
BaSO (s) Ba (aq)+SO (aq) K =[Ba ][SO ]
4 H O 2+
4 2
sp 2+
4 2
2 -
-
= (s) (s) K Ssp
2=
Ag CrO (s) 2Ag (aq)+CrO (aq)2 4
H O + 4
22 - →←
K =[Ag ] [CrO ] = (2s) (s)
K =4s
sp + 2
4 2
2
sp 3
-
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Summary
„ According to Arrhenius, an acid is a substance that dissociates to give hydrogen ions in water.
„ According to Lowry and Bronsted concept, an acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton from other substance.
„ According to Gilbert . N. Lewis , an acid is a species that accepts an electron pair while base is a species that donates an electron pair.
„ ionic product (ionic product constant) of water (Kw)=[H3O
+][OH–]
„ pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.
pH = –log 10 [H3O +]
„ when dilution increases, the degree of dissociation of weak electrolyte also increases. This statement is known as Ostwald’s dilution Law.
„ When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further this is known as common ion effect
„ Buffer is a solution which consists of a mixture of a weak acid and its conjugate base (or) a weak base and its conjugate acid.
„ Buffer capacity and buffer index is defined as the number of gram equivalents of acid or base added to 1
litre of the buffer solution to change its pH by unity.
β= dB d(pH)
„ Henderson – Hasselbalch equation
For Acid buffer
⇒ + pH = pK log [salt] [acid]a
For Basic buffer
⇒ +pOH = pK log [salt] [base]b
„ Hydrolysis of Salt of strong base and weak acid
K .K =Kh a w
pH = 7 + 1 2 pK log C.a + 1
2 „ Hydrolysis of salt of strong acid and
weak base
K .K =Kh b w
pH = 7 - 1 2 pK 1
2 log C.b −
„ Hydrolysis of Salt of weak acid and weak base
K .K .K =Ka b h w
pH = 7 + 1 2 pK pKa b−1
2 .
„ The solubility product of a compound is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co – efficient in a balanced equilibrium equation.
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Choose the correct answer:
1. Concentration of the Ag+ ions in a saturated solution of Ag C O2 2 4 is 2.24 10 mol L-4 -1× solubility product of Ag C O2 2 4 is (NEET – 2017)
a) 2.42 10-8× mol3L-3 b) 2.66 10-12× mol3L-3
c) 4.5 10-11× mol3L-3 d) 5.619 × 10–12 mol3L-3
2. Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations. (NEET – 2018)
i. 60 mL M 10
HCl + 40mL M 10
NaOH ii. 55 mL M 10
HCl + 45 mL M 10
NaOH
iii. 75 mL M 5
HCl + 25mL M 5
NaOH iv. 100 mL M 10
HCl + 100 mL M 10
NaOH
pH of which one of them will be equal to 1?
a) iv b) i c) ii d) iii
3. The solubility of BaSO4 in water is 2.42 10 gL-3 -1× at 298K. The value of its solubility product Ksp( ) will be (NEET -2018). (Given molar mass of BaSO =233g mol4
-1 )
a) 1.08 10 mol L-14 2 -2× b)1.08 10 mol L-12 2 -2×
c) 1.08 10 mol L-10 2 -2× d) 1.08 10 mol L-8 2 -2×
4. pH of a saturated solution of Ca(OH)2 is 9. The Solubility product ( )spK of Ca(OH)2
a) 0.5 10-15× b) 0.25 10-10×
c) 0.125 10-15× d) 0.5 10-10×
5. Conjugate base for Bronsted acids H O2 and HF are
a) - + 2OH and H FH , respectively b) H O and F , respectively3
+ -
c) OH and F , respectively- - d) H O and H F , respectively3 2 ++
6. Which will make basic buffer?
a) 50 mL of 0.1M NaOH+25mL of 0.1M CH COOH3
b) 100 mL of 0.1M CH COOH+100 mL of 0.1M NH OH3 4
c) 100 mL of 0.1M HCl+200 mL of 0.1M NH OH4
d) 100 mL of 0.1M HCl+100 mL of 0.1M NaOH
EVALUATION
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7. Which of the following fluro compounds is most likely to behave as a Lewis base? (NEET – 2016) a) BF3 b) PF3 c) CF4 d) SiF4
8. Which of these is not likely to act as Lewis base? a) BF3 b) PF3 c) CO d) F–
9. The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively
a) acidic, acidic, basic b) basic, acidic, basic
c) basic, neutral, basic d) none of these
10. The percentage of pyridine (C H N)5 5 that forms pyridinium ion (C H NH)5 5 in a 0.10M aqueous pyridine solution -9
b 5 5(K for C H N= 1.7 10 )× is
a) 0.006% b) 0.013% c) 0.77% d) 1.6%
11. Equal volumes of three acid solutions of pH 1,2 and 3 are mixed in a vessel. What will be the H+ ion concentration in the mixture? a) 3.7 10-2× b) 10-6 c) 0.111 d) none of these
12. The solubility of AgCl (s) with solubility product 1.6 10-10× in 0.1M NaCl solution would be
a) 1.26 10 M-5× b) 1.6 10 M-9× c) 1.6 10 M-11× d) Zero
13. If the solubility product of lead iodide is 3.2 10-8× , its solubility will be
a) -32×10 M b) 4 10 M-4× c) 1.6 10 M-5× d) 1.8 10 M-5×
14. MY and NY3 , are insoluble salts and have the same Ksp values of 6.2 10-13× at room temperature. Which statement would be true with regard to MY and NY ?3
a) The salts MY and NY3 are more soluble in 0.5M KY than in pure water b) The addition of the salt of KY to the suspension of MY and NY3 will have no effect
on their solubility’s c) The molar solubilities of MY and NY3 in water are identical d) The molar solubility of MY in water is less than that of NY3
15. What is the pH of the resulting solution when equal volumes of 0.1M NaOH and 0.01M HCl are mixed?
a) 2.0 b) 3 c) 7.0 d) 12.65
16. The dissociation constant of a weak acid is 1 10-3× . In order to prepare a buffer solution
with a pH = 4, the [Acid] [Salt]
ratio should be
a) 4:3 b) 3:4 c) 10:1 d) 1:10
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17. The pH of 10 M KOH-5 solution will be
a) 9 b) 5 c) 19 d) none of these
18. H PO2 4 - the conjugate base of
a) PO4 3− b) P O2 5 c) 3 4H PO d) HPO
4
2-
19. Which of the following can act as Lowry – Bronsted acid as well as base?
a) HCl b) SO4 2− c) HPO4
2− d) Br-
20. The pH of an aqueous solution is Zero. The solution is
a) slightly acidic b) strongly acidic c) neutral d) basic
21. The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by
a) [H ]= K [acid]
[salt] + a b) [H ]= K [salt]+
a c) [H ]=K [acid]+ a
d) [H ]= K [salt] [acid]
+ a
22. Which of the following relation is correct for degree of hydrolysis of ammonium acetate?
a) h = K C
h b) h = K K
a
b
c) w
a b
K h =
K .K d) h =
K .K K a b
w
23. Dissociation constant of NH OH4 is 1.8 10 -5× the hydrolysis constant of NH Cl4 would
be
a) 1.8 10-19× b) -105.55 10× c) 5.55 10-5× d) 1.80 10-5×
Answer the following questions: 1. What are Lewis acids and bases? Give two example for each.
2. Discuss the Lowry – Bronsted concept of acids and bases.
3. Indentify the conjugate acid base pair for the following reaction in aqueous solution
i)HS (aq) + HF F (aq) + H S(aq) ii) HPO + SO- - 2 4
2- 3
2- 4
3- 3
-
4 +
3 2-
3 3 -
PO + HSO iii)NH + CO NH + HCO
4. Account for the acidic nature of HClO4 in terms of Bronsted – Lowry theory, identify its conjugate base.
5. When aqueous ammonia is added to CuSO4 solution, the solution turns deep blue due to the formation of tetramminecopper (II) complex, [Cu(H O) ] +4NH (aq) [Cu(NH ) ]2 4 (aq)
2+ 3 3 4 (aq)
2+ , among H O2 and NH3 Which is stronger
Lewis base. 6. The concentration of hydroxide ion in a water sample is found to be 2.5 10 M-6× . Identify
the nature of the solution.
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7. A lab assistant prepared a solution by adding a calculated quantity of HCl gas at 25 Co to get a solution with [H O ]= 4 10 M3
+ -5× . Is the solution neutral (or) acidic (or) basic. 8. Calculate the pH of 0.04 M HNO3 Solution. 9. Define solubility product 10. Define ionic product of water. Give its value at room temperature. 11. Explain common ion effect with an example 12. Derive an expression for Ostwald’s dilution law
13. Define pH
14. Calculate the pH of -31.5×10 M solution of Ba (OH) 2
15. 50ml of 0.05M HNO3 is added to 50ml of 0.025M KOH . Calculate the pH of the resultant solution.
16. The K a value for HCN is 10-9 . What is the pH of 0.4M HCN solution?
17. Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that K =K =1.8 10a b
-5×
18. Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base
19. Solubility product of Ag CrO2 4 is 1 10-12× . What is the solubility of Ag CrO2 4 in 0.01M AgNO3 solution?
20. Write the expression for the solubility product of Ca (PO )3 4 2
21. A saturated solution, prepared by dissolving 2CaF (s) in water, has [Ca ]=3.3 10 M 2+ -4× What
is the Ksp of CaF2 ?
22. Ksp of AgCl is 1 8 10 10. × − . Calculate molar solubility in 1 M AgNO3
23. A particular saturated solution of silver chromate 2 4Ag CrO has [Ag ]=5 10+ -5× and [CrO ] =4.4 10 M.4
2 -4- × What is the value of Ksp for Ag CrO2 4 ?
24. Write the expression for the solubility product of Hg Cl .2 2
25. Ksp of Ag CrO2 4 is 1.1 10 -12× . what is solubility of Ag CrO
2 4 in 0.1M K CrO2 4 .
26. Will a precipitate be formed when 0.150 L of 0.1M 3 2Pb(NO ) and 0.100L of 0.2 M NaCl are mixed? K (PbCl )=1.2 10 .sp 2
-5×
27. Ksp of Al(OH)3 is 1 10 M-15× . At what pH does 1.0 10 M Al-3 3+× precipitate on the addition of buffer of NH Cl4 and NH OH4 solution?
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Buffers and pH
Step – 1 Open the Browser and type the URL given (or) Scan the QR Code. You can see a webpage as shown in the
figure.
Step – 2 Now you can select a combination of an acid/base (Box 1) and its corresponding salt (Box 2) from the given
choices and also select the desired concentrations (Box 3) and volume (Box 4) of these for the buffer.
Step – 3 In order to measure the pH of the made-up buffer click the ‘Insert Probe’ (Box 5) on the pH meter. Now the
pH meter shows the pH. After measuring you need to remove the probe by clicking ‘Remove Probe” (Box 5)
to make any changes in the composition.
Step – 4 Now you can vary the concentration and volume of the components and see how the pH changes.
By using this tool you can simulate the prepara� on of a buff er and measure its pH values
Please go to the URL h� p://pages.uoregon.edu/ tgreenbo/pHbuff er20.html (or) Scan the QR code on the right side
ICT Corner
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