Numerical Problems with Answers, Solution

Example 8.1

Calculate the concentration of OH- in a fruit juice which contains 2 ×10-3 M, H3O+ ion. Identify the nature of the solution.

Given that H3O+ = 2 ×10-3 M

Kw = [H3O+ ][OH - ]

∴[OH - ]= Kw /[H3O+ ] = [1 ××10-14] / [2 x 10 3] =5 ×10-12 M

 2 ×10-3 >> 5 ×10-12

 i.e., [H3O+ ] >> [OH-], hence the juice is acidic in nature

Example 8.2

Calculate the pH of 0.001M HCl solution

H3O+ from the auto ionisation of H2O (10-7M) is negligible when compared to the H3O+ from 10-3M HCl.

Hence [H3O+ ]= 0.001 mol dm–3

pH = -log10 [H3O+]

= -log10 (0.001)

= -log10 (10-3 ) = 3

Note: If the concentration of the acid or base is less than 10–6 M, the concentration of H3O+ produced due to the auto ionisation of water cannot be negleted and in such cases

[H3O+] = 10-7(from water) + [H3O+] (from the acid)

similarly, [OH-] = 10-7 M (from water) + [OH-] (from the base)

Example 8.3

Calculate pH of 10-7 M HCl

If we do not consider [H3O+ ] from the ionisation of H2O,

then [H3O+ ] = [HCl] = 10-7M

i.e., pH = 7, which is a pH of a neutral solution. We know that HCl solution is acidic whatever may be the concentration of HCl i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10-7M) Hence, the H3O+ (10-7M) formed due to the auto ionisation of water cannot be neglected.

so, in this case we should consider [H3O+ ] from ionisation of H 2O

[H3O+ ] = 10-7 (from HCl) + 10-7 (from water)

= 10-7 (1+1)

= 2 ×10-7

pH = -log10[H3O+ ]

= - log 2 - (-7).log10 10

= 7-log 2

= 7-0.3010 = 6.6990

= 6.70

Example 8.4

A solution of 0.10M of a weak electrolyte is found to be dissociated to the extent of 1.20% at 25oC . Find the dissociation constant of the acid.

Given that α=1.20%=1.20/100

= 1.2 × 10-2

Ka = α2c

= (1.2 × 10-2 )2 (0.1) = 1.44 ×10-4 ×10-1

= 1.44 × 10-5

Example 8.5

Calculate the pH of 0.1M CH3COOH solution. Dissociation constant of acetic acid is 1.8 ×10-5 .

pH=-log[H+ ]

For weak acids,

[H+ ]= √ [Ka × C]

= √ (1.8 ×10-5 × 0.1)

=1.34 ×10-3M      pH=-log (1.34 ×10-3)

= 3 - log1.34

= 3 - 0.1271

= 2.8729 ≈2.87

Example 8.6

Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per litre acetic acid. Ka for acetic acid is 1.8 × 10-5 .

pH = pKa + log {[salt]/[acid]}

Given that Ka = 1.8 × 10-5

∴ pKa = −log(1.8 ×10−5 ) = 5 −log1.8

= 5 -0.26

= 4.74

∴pH = 4.74 + log (0.20/0.18)

= 4.74 + log (10/9) = 4.74 + log 10 - log 9

= 4.74 + 1 - 0.95 = 5.74 - 0.95

= 4.79

Example 8.7

What is the pH of an aqueous solution obtained by mixing 6 gram of acetic acid and 8.2 gram of sodium acetate making the volume equal to 500 ml. (Given: Ka for acetic acid is 1.8 × 10-5 )

According to Henderson – Hasselbalch equation,

pH = pKa + log {[salt]/[acid]}

PKa =-logKa =-log(1.8 ×10-5 )= 4.74 (Refer previous example)

Salt = Number of moles of sodium acetate / Volume of the solution ( litre )

Number of moles of sodium acetate = mass of sodium acetate / molar mass of sodium acetate

pH = 4.74 + log 1

pH = 4.74 + 0 = 4.74

Example 8.8

Calculate i) degree of hydrolysis, ii) the hydrolysis constant and iii) pH of 0.1M CH3COONa solution ( pKa for CH3COOH is 4.74).

Solution (a) CH3COONa is a salt of weak acid (CH 3COOH) and a strong base (NaOH).

Hence, the solutions is alkaline due to hydrolysis.

CH3COO - (aq) + H2O(aq) ↔ CH3COOH (aq) + OH-(aq)

Book Back Questions:

8. Calculate the pH of 0.04 M HNO3 Solution.

Solution:

Concentration of HNO3 = 0.04M

[H3O +] = 0.04 mol dm-3

pH = -log[H3O+]

 = -log(0.04)

 = -log(4 ×10-2 )

= 2-log4

=2-0.6021

=1.3979 = 1.40

14. Calculate the pH of 1.5×10-3M solution of Ba (OH)2

Answer:

Ba(OH)2 → Ba2+ + 2OH-

1.5 ×10-3M

2 ×1.5 × 10-3M.

[OH-] = 3 ×10-3M

[ pH+pOH=14]

pH= 14-pOH

pH= 14- ( −log [OH- ] )

= 14+ log [OH- ]

=14+log(3 ×10-3 )

= 14 + log 3 + log 10-3

= 14 + 0.4771−3

= 11+0.4771

 pH=11.48

15. 50ml of 0.05M HNO3 is added to 50ml of 0.025M KOH . Calculate the pH of the resultant solution.

Answer:

Number of moles of HNO3 = 0.05 × 50 × 10-3

= 2.5 × 10-3

Number of moles of KOH = 0.025 × 50 × 10−3

= 1.25 × 10-3

Number of moles of HNO3 after mixing

 = 2.5 × 10-3– 1.5 × 10-3

= 1.25 × 10-3

∴ concentration of HNO3 = Number of moles of HNO3 / Volume is litre

After mixing, total volume = 100 ml = 100×10-3L

∴[H + ]= 1.25 ×10-3 moles / 100 ×10-3L

= 1.25 ×10-2 moles L-1

pH = - log [H+ ]

pH = − log(1.25 ×10−2 )

= 2 − 0.0969

= 1.9031

16. The Ka value for HCN is 10-9 . What is the pH of 0.4M HCN solution?

Answer:

Given

Ka =10-9

c=0.4M

pH=-log[H+ ]

 [H+ ] = √ (Ka × c.)

 = √(10-9 ×0.4)

 = 2 ×10-5

∴pH = - log (2 ×10-5 )

=5-log2

= 5-0.3010

=4.699.

17. Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that Ka =Kb =1.8 ×10-5

Answer:

 h = √Kh = √ [Kw / KaKb]

= √ { 1×10−14  /  1.8×10−5 × 1.8 ×10−5 }

= √ {1/1.8 ×10−4}

= 0.7453 ×10−2

pH = 1/2 pKw + 1/2 pKa - 1/2 pKb

Given that Ka = Kb = 1.8 × 10-5

if Ka = Kb , then, pKa = pKb

∴ pH = 1/2 pKw = 1/2 (14) = 7

19. Solubility product of Ag2CrO4 is 1 ×10-12 . What is the solubility of Ag2 CrO4 in 0.01M AgNO3 solution?

Answer:

Given that Ksp =1 × 10-12

Ag2 CrO4 (s) ↔ 2 Ag+(aq) + CrO4−2 (aq)

 s ↔ 2s + s

AgNO3 (s) ↔ Ag+(aq) + NO3- (aq)

 0.01M ↔ 0.01M + 0.01M

[Ag+] = 2s + 0.01     

0.01>>2S 

∴ [Ag+ ]= 0.01M

[CrO42−]= s

Ksp = [Ag+]2 [CrO42- ]

1 × 10-12 = (0.01)2 (s)

(s) = 1 × 10-12 / (10-2)2 = 1 × 10−8 M

20. Write the expression for the solubility product of Ca3 (PO4 )2

Answer:

Ca3 (PO4 )2 s ↔ 3Ca2+ 3s + 2PO43- 2s

Ksp = [Ca2+]3 [PO43-]2

Ksp = (3s)3 (2s)2

Ksp = 27s3 .4s2

Ksp =108s5

21. A saturated solution, prepared by dissolving CaF2 (s) in water, has [Ca2+] = 3.3 ×10-4M What is the Ksp of CaF2 ?

Answer:

CaF2(s) ↔ Ca(aq)2+ + 2F -(aq)

[F -] = 2 [Ca2+ ] = 2 × 3.3 ×10-4 M

 = 6.6 ×10-4M

Ksp = [Ca2+][F -]2

= (3.3 ×10-4)(6.6 ×10-4)2

= 1.44 ×10-10

22. Ksp of AgCl is 1.8 ×10−10 . Calculate molar solubility in 1 M AgNO3

Answer:

AgCl (s) ↔ Ag+ (aq) + Cl- (aq)

x = solubility of AgCl in 1M AgNO3

AgNO3 (aq) ↔ Ag+ (aq) 1M + NO3-(aq) 1M

 [Ag+] = x+1 = 1M           (x<<1)

 [Cl-] = x

Ksp =[Ag+][Cl-]

1.8 × 10-10 = (1)(x)

x= 1.8 × 10-10 M

23. A particular saturated solution of silver chromate Ag2 CrO4 has [Ag+ ]=5 ×10-5 and [CrO4 ]2 -=4.4 ×10 -4 M. What is the value of Ksp for Ag2 CrO4 ?

Answer:

Ag2 CrO4 (s) ↔ 2Ag+(aq) + CrO42- (aq)

Ksp = [Ag+]2[CrO42-]

 = (5×10-5)2 (4.4×10 -4 )

 = 1.1×10-12

24. Write the expression for the solubility product of Hg2 Cl2 .

Answer:

Hg2 Cl2 s ↔ Hg22+ + 2Cl- 2s

Ksp = [Hg22+][Cl-]2

= (s)(2s)2

Ksp =4s3

25. Ksp of Ag2 CrO4 is 1.1×10-12 . what is solubility of Ag2 CrO4 in 0.1M K2 CrO4.

Answer:

Ag2CrO4 x ↔ 2Ag+ 2x + CrO42-  x

x is the solubility

of Ag2 CrO4 in 0.1M K2CrO4

K2CrO4 ↔ 2K+   + CrO42-

 0.1M ↔ 0.2 M + 0.1 M

[Ag+] = 2x

 [CrO42- ] = (x+0.1) ≈ 0.1           x<<0.1

Ksp = [Ag+]2[CrO42-]

 1.1 ×10-12 = (2x)2 (0.1)

1.1 ×10-12 = 0.4x2

x2 = [ 1.1 ×10-12 / 0.4]

x = √[ 1.1 ×10-12 / 0.4]

x= √2.75 ×10-12

x=1.65 ×10-6M 

26. Will a precipitate be formed when 0.150 L of 0.1M Pb(NO3)2 and 0.100L of 0.2 M NaCl are mixed? Ksp (PbCl2 )=1.2 ×10-5 .

Answer:

When two are more solution are mixed, the resulting concentrations are differnet from the original.

Total volume = 0.250L

Pb(NO3)2 ↔ Pb2+ + 2NO3-

0.1 M  ↔ 0.1 M + 0.2 M

Number of moles

Pb2+ = molarity × Volume of the solution in lit

= 0.1 × 0.15

[Pb2+ ]mix = 0.1 × 0.15 /  0.25

= 0.06 M

NaCl ↔ Na+ + Cl-

0.2 M ↔ 0.2 M + 0.2 M

No.of moles Cl- = 0.2 × 0.1

[Cl-]mix = (0.2 × 0.1) / 0.25  = 0.08 M

Precipitation of PbCl2 (s) occurs if

[Pb2+ ][Cl- ]2 > Ksp

[Pb2+ ][Cl- ]2 =(0.06)(0.08)2

 = 3.84 ×10-4

Since ionic product [Pb2+ ][Cl- ]2 >Ksp ,

PbCl2 is precipitated

27. Ksp of Al(OH)3 is 1 ×10-15M . At what pH does 1.0 ×10-3M Al3+ precipitate on the addition of buffer of NH4Cl and NH4OH solution?

Answer:

Al(OH)3 ↔ Al 3+ (aq)+3OH- (aq)

Ksp = [Al3+][OH-]3

Al(OH)3 precipitates when

[Al3+][OH-]3 >Ksp

(1×10-3)[OH-]3 > 1×10-15

[OH]3 >1×10-12

[OH-] > 1×10-4M

[OH- ] = 1×10-4M

POH = - log10[OH-] = -log(1 ×10-4 )=4

pH = 14-4= 10

Thus, Al (OH)3 precipitates at a pH of 10