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Solutions/Answers to NCERT/CBSE CHEMISTRY Class 11(Class XI)textbook ... 7.73 The concentration of sulphide ion in 0.1M HCl solution saturated with.
NCERT/CBSE CHEMISTRY CLASS 11 textbook http://www.TutorBreeze.com Contact for Online Tutoring in Physics, Math, Chemistry, English, Accounts, MBA Solutions/Answers to NCERT/CBSE CHEMISTRY Class 11(Class XI)textbook CHAPTER SEVEN EQUILIBRIUM 7.73 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place? Solution:

Precipitation takes place in the solution when: Ionic Product>Solubility Product Let the metal ions be denoted by M.(All have equal concentrations) The reactions is: MS  M 2+ + S 2− 5 × 0.04 [ M 2+ ] = = 0.0133 15 10 [S2- ] = 1.0 × 10 −19 × = 6.67 ×10 −20 15 2+ Ionic Product=[M ][ S 2 − ] = 0.0133 × 6.67 ×10 −20 = 8.87 × 10−22 Comparing with the given values, ZnCl2 and CdCl2 solutions will be precipitated.

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NCERT/CBSE CHEMISTRY CLASS 11 textbook http://www.TutorBreeze.com Contact for Online Tutoring in Physics, Math, Chemistry, English, Accounts, MBA .70 The ionization constant of benzoic acid is 6.46 × 10–5 and K for silver benzoate sp is 2.5 × 10–13. How many times is silver benzoate more

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Let the solubility of silver benzoate in water be s. The reaction is: C6 H 5COOAg  C6 H 5COO −1 +Ag + At eqbm.:

s

s

−1

+

We have,K SP = [C6 H 5COO ][Ag ] ⇒ 2.5 × 10−13 = s 2 Solving ,s=5.0 × 10-7 M Now in case of buffer of given pH, 3.19=-log[H + ] ⇒ [ H + ] = 10 −3.19 = Antilog(4.81)=6.457 ×10-4 Reaction for dissociation of benzoic acid is: C6 H 5COOH  C6 H 5COO −1 +H + Concentration of hydrogen ions can be assumed to be constant as the solution is a buffer. Assume solubility in buffer solution to be s'. We have, Ka = ⇒

[C6 H 5COO −1 ][H + ] = 6.46 × 10−5 [C6 H 5 COOH] [C6 H 5 COOH] 6.457 ×10-4 = ≈ 10 [C6 H 5COO −1 ] 6.46 × 10−5 Now:

[Ag]+ = s ' = [C6 H 5COO −1 ] + [C6 H 5COOH ](These values are to be substituted from above). ⇒ s'=11[C6 H 5COO −1 ] s' 11 C6 H 5COOAg  C6 H 5COO −1 +Ag + ⇒ [C6 H 5COO −1 ] =

At eqbm.:

s' 11

s'

s '2 = 1.66 × 10 −6 11 s' 1.66 × 10−6 Now ratio of solubilities= = = 3.32 s 5.0 × 10-7 Thus,silver benzoate is 3.32 times more soluble in a buffer of pH 3.19 ©TutorBreeze.com compared to its solubility in pure water. Now,k SP = [C6 H 5COO −1 ][Ag + ] =

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