Chapter 7 Equilibrium Solutions
Question - 41 : - The concentration of hydrogen ion in asample of soft drink is 3.8 × 10–3 M. what is its pH?
Answer - 41 : -
Given,
pH value of soft drink
Question - 42 : - ThepH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ionin it.
Answer - 42 : -
Given,
pH= 3.76
Itis known that,
Hence,the concentration of hydrogen ion in the given sample of vinegar is 1.74 × 10–4 M.
Question - 43 : - The ionization constant of HF, HCOOH and HCNat 298K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively.Calculate the ionization constants of the corresponding conjugate base.
Answer - 43 : -
Itis known that,
Given,
Ka of HF = 6.8 × 10–4
Hence, Kb of its conjugatebase F–
Given,
Ka of HCOOH = 1.8 × 10–4
Hence, Kb of its conjugatebase HCOO–
Given,
Ka of HCN = 4.8 × 10–9
Hence, Kb of its conjugatebase CN–
Question - 44 : - The ionization constant of phenol is 1.0 ×10–10. What is the concentration of phenolate ionin 0.05 M solution of phenol? What will be its degree of ionization if thesolution is also 0.01M in sodium phenolate?
Answer - 44 : -
Ionizationof phenol:
Now, let ∝ be the degree of ionizationof phenol in the presence of 0.01 M C6H5ONa.
Also,
Question - 45 : - The first ionization constant of H2S is 9.1 × 10–8. Calculate theconcentration of HS– ion in its 0.1 M solution. How willthis concentration be affected if the solution is 0.1 M in HCl also? If thesecond dissociation constant of H2S is 1.2 × 10–13, calculate theconcentration of S2– under both conditions.
Answer - 45 : -
(i) To calculate theconcentration of HS– ion:
Case I (in theabsence of HCl):
Let the concentration of HS– be x M.
Case II (in thepresence of HCl):
Inthe presence of 0.1 M of HCl, let 
be y M.
(ii) Tocalculate the concentration of
:
Case I (in theabsence of 0.1 M HCl):
(Fromfirst ionization, case I)
Let 
Also,
(Fromfirst ionization, case I)Case II (in thepresence of 0.1 M HCl):
Again, let the concentration of HS– be X’ M.
(Fromfirst ionization, case II)
(FromHCl, case II)
Question - 46 : - The ionization constant of acetic acid is1.74 × 10–5. Calculate the degree of dissociation ofacetic acid in its 0.05 M solution. Calculate the concentration of acetate ionin the solution and its pH.
Answer - 46 : -
Method 1
Since Ka >> Kw, :
Method 2
Degreeof dissociation,
c = 0.05 M
Ka = 1.74 × 10–5
Thus, concentration of CH3COO– = c.α
Hence,the concentration of acetate ion in the solution is 0.00093 M and its Ph is3.03.
Question - 47 : - It has been found that the pH of a 0.01Msolution of an organic acid is 4.15. Calculate the concentration of the anion,the ionization constant of the acid and its pKa.
Answer - 47 : -
Letthe organic acid be HA.
Concentrationof HA = 0.01 M
pH= 4.15
Now, 
Then,
Question - 48 : - Assumingcomplete dissociation, calculate the pH of the following solutions:
(a)0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH
Answer - 48 : -
(i) 0.003MHCl:
SinceHCl is completely ionized,
Now,
Hence, the pH of the solution is 2.52.
(ii) 0.005MNaOH:
Hence,the pH of the solution is 11.70.
(iii) 0.002 HBr:
Hence,the pH of the solution is 2.69.
(iv) 0.002 M KOH:
Hence,the pH of the solution is 11.31.
Question - 49 : - Calculatethe pH of the following solutions:
a)2 g of TlOH dissolved in water to give 2 litre of solution.
b) 0.3 g of Ca(OH)2 dissolved in water to give500 mL of solution.
c)0.3 g of NaOH dissolved in water to give 200 mL of solution.
d)1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.
Answer - 49 : -
(a) For2g of TlOH dissolved in water to give 2 L of solution:
(b) For 0.3 g of Ca(OH)2 dissolved in waterto give 500 mL of solution:
(c) For 0.3 g of NaOHdissolved in water to give 200 mL of solution:
(d) For 1mL of 13.6 MHCl diluted with water to give 1 L of solution:
13.6 × 1 mL = M2 × 1000 mL
(Beforedilution) (After dilution)
13.6 × 10–3 = M2 × 1L
M2 = 1.36 × 10–2
[H+] = 1.36 × 10–2
pH = – log (1.36 × 10–2)
=(– 0.1335 + 2)
=1.866 ∼ 1.87
Question - 50 : - The degree of ionization of a 0.1Mbromoacetic acid solution is 0.132. Calculate the pH of the solution andthe pKa of bromoacetic acid.
Answer - 50 : -
Degreeof ionization, α = 0.132
Concentration, c = 0.1 M
Thus, the concentration of H3O+ = c.α
=0.1 × 0.132
=0.0132
Now,
