Chapter 7 Equilibrium Solutions
Question - 61 : - The ionization constant of nitrous acid is4.5 × 10–4. Calculate the pH of 0.04 M sodium nitritesolution and also its degree of hydrolysis.
Answer - 61 : -
NaNO2 is the salt of astrong base (NaOH) and a weak acid (HNO2).
Now, If x moles of the saltundergo hydrolysis, then the concentration of various species present in thesolution will be:
Therefore,degree of hydrolysis
=2.325 × 10–5
Question - 62 : - A0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate theionization constant of pyridine
Answer - 62 : -
pH= 3.44
Weknow that,
pH = – log [H+]
Question - 63 : - Predictif the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF
Answer - 63 : -
(i) NaCl:
Therefore,it is a neutral solution.
(ii) KBr:
Therefore,it is a neutral solution.
(iii) NaCN:
Therefore,it is a basic solution.
(iv) NH4NO3
.
Therefore,it is an acidic solution.
(v) NaNO2
Therefore,it is a basic solution.
(vi) KF
Therefore,it is a basic solution.
Question - 64 : - The ionization constant of chloroacetic acidis 1.35 × 10–3. What will be the pH of 0.1M acid and its0.1M sodium salt solution?
Answer - 64 : -
It is given that Ka for ClCH2COOH is 1.35 × 10–3.
ClCH2COONa is the salt of aweak acid i.e., ClCH2COOH and a strong base i.e., NaOH.
Question - 65 : - Ionic product of water at 310 K is 2.7 × 10–14. What is the pH ofneutral water at this temperature?
Answer - 65 : -
Ionicproduct,
Hence,the pH of neutral water is 6.78.
Question - 66 : - Calculatethe pH of the resultant mixtures:
a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01MCa(OH)2
c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
Answer - 66 : -
(a) 
Thus, excess of 
=.0015 mol
(b) 
Since there is neither an excess of 
or,the solution is neutral. Hence, pH = 7.
(c) 
Excess of =.001 mol.
Question - 67 : - Determinethe solubilities of silver chromate, barium chromate, ferric hydroxide, leadchloride and mercurous iodide at 298K from their solubility product constantsgiven in Table 7.9 (page 221). Determine also the molarities of individualions.
Answer - 67 : -
(1) Silverchromate:
Let the solubility of 
bes.
Molarity of 
= 2s = 2 × 0.65 × 10–4 = 1.30 × 10–4 M
Molarityof 
= s = 0.65 × 10–4 M(2) Bariumchromate:
Let s be the solubilityof 
Thus, 
= s and 
= s
Molarity of 
= Molarity of 
(3) Ferrichydroxide:
Let s be the solubilityof 
Molarity of 
Molarity of 
(4) Lead chloride:
Let KSP be the solubility of 
Molarity of 
Molarity of chloride =
.
(5) Mercurousiodide:
Let s be the solubility of 
Molarity of 
Molarity of 
Question - 68 : - The solubility product constant of Ag2CrO4 and AgBr are 1.1 ×10–12 and 5.0 × 10–13 respectively.Calculate the ratio of the molarities of their saturated solutions.
Answer - 68 : -
Let s be the solubility ofAg2CrO4.
Let s´ be the solubility ofAgBr.
Therefore,the ratio of the molarities of their saturated solution is 
Question - 69 : - Equal volumes of 0.002 M solutions of sodiumiodate and cupric chlorate are mixed together. Will it lead to precipitation ofcopper iodate? (For cupric iodate Ksp = 7.4 × 10–8).
Answer - 69 : -
Whenequal volumes of sodium iodate and cupric chlorate solutions are mixedtogether, then the molar concentrations of both solutions are reduced to halfi.e., 0.001 M.
Then,
Now,the solubility equilibrium for copper iodate can be written as:
Ionicproduct of copper iodate:
Sincethe ionic product (1 × 10–9) is less than Ksp (7.4 × 10–8), precipitation will notoccur.
Question - 70 : - The ionization constant of benzoic acid is6.46 × 10–5 and Ksp for silver benzoateis 2.5 × 10–13. How many times is silver benzoate moresoluble in a buffer of pH 3.19 compared to its solubility in pure water?
Answer - 70 : -
SincepH = 3.19,
Let the solubility of C6H5COOAg be x mol/L.
Then,
Thus, the solubility of silver benzoate in apH 3.19 solution is 1.66 × 10–6 mol/L.
Now, let the solubility of C6H5COOAg be x’ mol/L.
Hence,C6H5COOAg is approximately3.317 times more soluble in a low pH solution.