Ncert Solutions Chemistry Class 11th
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7.66. Calculate the 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 0.01M Ca(OH)2
c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
7.66. Calculate the 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 0.01M Ca(OH)2
c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
(a) Total number of moles present in 10 mL of 0.2 M calcium hydroxide are (10×0.2) / 1000? = 0.002 moles.
Total number of moles present in 25 mL of 0.1 M HCl are (25×2) / 1000? = 0.0025 moles.
Ca (OH)2? + 2HCl → CaCl2? + 2H2? O
1 mole of calcium hydroxide reacts with 2 moles of HCl.
0.0025
Upvote 
7.65. Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
7.65. Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
Ionic product,
Kw= [H+] [OH−]
Let [H+]= x
Since [H+]= [OH−], Kw=x2.
⇒Kw at 310K is 2.7×10−14
∴2.7×10−14=x2
⇒x=1.64×10−7
⇒ [H+]=1.64×10−7
⇒ pH= −log [H+]=−log [1.64×10−7]=6.78
Hence, the pH of neutral water is 6.78.
Upvote 7.64. The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
7.64. The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
Given Ka=1.35 x 10−3
For acid solution:
[H+] = (KaC)1/2 = (0.00135 x 0.1)1/2 = 0.0116M
pH = – log [H+] = –log0.0116 = 1.936
For 0.1M sodium salt solution
ClCH2COONa is the salt of a weak acid i.e., ClCH2COOH and a strong base i.e., NaOH.
pKa = -logKa = log (0.00135) = 2.8697
pKw = 14
logc = lo
Upvote 7.63. Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH4NO3, NaNO2 and
7.63. Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH4NO3, NaNO2 and
(i) NaCl:
NaCl+H2O↔NaOH+HCl
Strong base Strong acid
Therefore, it is a neutral solution.
(ii) KBr:
KBr+H2O↔KOH+HBr
Strong base Strong acid
Therefore, it is a neutral solution.
(iii) NaCN:
NaCN+H2O↔HCN+NaOH
Weak acid Strong base
Therefore, it is a basic solution.
(iv) NH4NO3
NH4N
Upvote 7.62. A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
7.62. A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
pH=3.44
We know that,
pH=−log [H+]
∴ [H+]=3.63×10−4
Then, Kb= (3.63×10−4)2 / 0.02 (? concentration =0.02M)
⇒Kb=6.6×10−6
Now, Kb=Kw / Ka
⇒Ka=Kw / Kb=10−14 / 6.6×10−6=1.51×10−9
Upvote 7.61. The ionization constant of nitrous acid is 4.5×10−4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
7.61. The ionization constant of nitrous acid is 4.5×10−4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Since NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2);
NO2− + H2O ↔ HNO2 + OH−
Then, Kb= [HNO2] [OH−] / [NO2−]
⇒ Kw/ Ka = (10−14) / (4.5×10−4)
= 0.22×10−10
Now, if x moles of the salt undergo hydrolysis, and then the concentration of various species present in the solution will be:
Upvote 7.60. The pHof 0.1M of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
7.60. The pHof 0.1M of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
c=0.1M
pH= −log [H+]
=> 2.34 = −log [H+]
So, −log [H+]= 2.34
=> [H+]= 4.5×10−3
Also,
[H+]= cα
=>4.5×10−3= 0.1×α
=> α=4.5×10−2= 0.045
Then,
Ka = α2/c = (45×10−3)2/ 0.1
=202.5×10−6
=2.02×10−4
Upvote 7.59. The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
7.59. The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
Let the degree of ionization of propanoic acid be α. Then, representing propionic acid as HA, we have:
HA + H2O ⇔ H3O+ + A−
(.05−0.0α)≈0.5
Upvote 7.58. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
7.58. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Solubility of Sr (OH)2=19.23g/L
The molecular weight of Sr (OH)2 is 87.6 + 2 (17)=121.6
Then, concentration of Sr (OH)2=19.23 /121.63M=0.1581M
Sr (OH)2 (aq)→Sr2+ (aq)+2 (OH−) (aq)
∴ [Sr2+]=0.1581M
[OH−]=2×0.1581M=0.3126M
Now, Kw= [OH−] [H+]
=> [H+] = 10−14 / 0.3126
=> [H+]=3.2×10−14
Upvote 7.57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
7.57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
[KOH]= [K+]= [OH−]= (0.561×1000) / (56×200)? =0.050M
[H+]=Kw / [OH−]? =10−14 / 0.05? =2.0×10−13
pH=−log [H+]=−log (2.0×10−13)
=12.7
Upvote