NCERT Class XI Chemistry Equilibrium Solutions
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Question : 66
Total: 73
Calculate the pH of the resultant mixtures :
(a) 10 mL of 0.2 MC a ( O H ) 2
(b) 10 mL of 0.01 MH 2 S O 4 C a ( O H ) 2
(c) 10 mL of 0.1 MH 2 S O 4
(a) 10 mL of 0.2 M
(b) 10 mL of 0.01 M
(c) 10 mL of 0.1 M
Solution:
(a) 10 mL of 0.2 M C a ( O H ) 2
= 2 millimoles ofC a ( O H ) 2
25 mL of 0.1 M HCl = 25 × 0.1 millimoles = 2.5 millimoles of HCl
C a ( O H ) 2 + 2 H C l C a C l 2 + 2 H 2 O
1 millimole ofC a ( O H ) 2
∴ 2.5 millimoles of HCl will react with 1.25 millimoles ofC a ( O H ) 2
∴C a ( O H ) 2
Total volume of the solution = 10 + 25 mL = 35 mL
∴ Molarity of Ca(OH)2 in the mixture solution =
∴[ O H – ] 10 – 2
pOH = –log(4.28 ×10 – 2
∴ pH = 14 – 1.37 = 12.63
(b) 10 mL of 0.01 MH 2 S O 4
10 mL of 0.01 MC a ( O H ) 2
C a ( O H ) 2 + H 2 S O 4 C a S O 4 + 2 H 2 O
1 mole ofC a ( O H ) 2 H 2 S O 4
∴ 0.1 millimole ofC a ( O H ) 2 H 2 S O 4
(c) 10 mL of 0.1 MH 2 S O 4
10 mL of 0.1 M KOH = 1 millimole
2 K O H + H 2 S O 4 K 2 S O 4 + 2 H 2 O
1 millimole of KOH will react with 0.5 millimole of H2SO4
∴H 2 S O 4
Volume of reaction mixture = 10 + 10 = 20 mL
∴ Molarity ofH 2 S O 4
10 − 2 M
[ H + ] 10 – 2 10 – 2
pH = –log(5 ×10 – 2
= 2 millimoles of
25 mL of 0.1 M HCl = 25 × 0.1 millimoles = 2.5 millimoles of HCl
1 millimole of
∴ 2.5 millimoles of HCl will react with 1.25 millimoles of
∴
Total volume of the solution = 10 + 25 mL = 35 mL
∴ Molarity of Ca(OH)2 in the mixture solution =
∴
pOH = –log(4.28 ×
∴ pH = 14 – 1.37 = 12.63
(b) 10 mL of 0.01 M
10 mL of 0.01 M
1 mole of
∴ 0.1 millimole of
(c) 10 mL of 0.1 M
10 mL of 0.1 M KOH = 1 millimole
1 millimole of KOH will react with 0.5 millimole of H2SO4
∴
Volume of reaction mixture = 10 + 10 = 20 mL
∴ Molarity of
pH = –log(5 ×
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