Explanation H 2 S is a diprotic protic acid. The primary transfer equilibrium is given below, H 2 S ( a q ) + H 2 O ( l ) ⇌ H 3 O + ( a q ) + H S − ( a q ) , K a 1 = 1.3 × 10 − 7 The expression for equilibrium constant for the first deprotonation reaction can be written as shown below, K a 1 = [ H 3 O + ] [ H S − ] [ H 2 S ] An equilibrium table can be set up as given below. C o m p o s i t i o n ( M ) H 2 S H 2 O H 3 O + H S − I n i t i a l c o n c e n t r a t i o n 0.20 − 0 0 C h a n g e i n c o n c e n t r a t i o n − x − + x + x E q u i l i b r i u m c o n c e n t r a t i o n 0.20 − x − x x Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below. K a 1 = ( x ) ( x ) ( 0.20 − x ) Now, the above expression can be solved for x. K a 1 = ( x ) ( x ) ( 0.20 − x ) 1.3 × 10 − 7 = ( x ) ( x ) ( 0.20 − x ) ( 1.3 × 10 − 7 ) ( 0.20 − x ) = x 2 x 2 + 1.3 × 10 − 7 x − 2.6 × 10 − 8 = 0 x = ( − 1.3 × 10 − 7 ) ± ( 1.3 × 10 − 7 ) 2 − ( 4 ) × ( 1 ) × ( − 2.6 × 10 − 8 ) 2 x = 1.6 × 10 − 4 [ H 3 O + ] = x = 1 . 6 × 1 0 - 4 M [ H S − ] = x = 1 . 6 × 1 0 - 4 M [ H 2 S ] = 0.20 − x ≈ 0 . 2 0 M Therefore, the concentration of H S − a n d H 3 O + is 1.6 × 10 − 4 M a n d 1.6 × 10 − 4 M respectively. The second deprotonation reaction is given below, H S − ( a q ) + H 2 O ( l ) ⇌ H 3 O + ( a q ) + S 2 − ( a q ) , K a 2 = 7.1 × 10 − 15 The expression for equilibrium constant for the second deprotonation reaction can be written as shown below, K a 2 = [ H 3 O + ] [ S 2 − ] [ H S − ] An equilibrium table can be set up as given below

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Chapter 6, Problem 6E.3AST

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The concentrations of all solute species in 0.20 M H2S(aq) have to be calculated.

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Chapter 6, Problem 6E.2BST

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Ch. 6 - Prob. 6A.1AST Ch. 6 - Prob. 6A.1BST Ch. 6 - Prob. 6A.2AST Ch. 6 - Prob. 6A.2BST Ch. 6 - Prob. 6A.3AST Ch. 6 - Prob. 6A.3BST Ch. 6 - Prob. 6A.1E Ch. 6 - Prob. 6A.2E Ch. 6 - Prob. 6A.3E Ch. 6 - Prob. 6A.4E

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The concentrations of all solute species in 0.20 M H 2 S ( a q ) have to be calculated.

Solution Summary: The author explains that the equilibrium constant for the first deprotonation reaction can be written as shown below.

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Interpretation:

The concentrations of all solute species in 0.20 M H2S(aq) have to be calculated.

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Chapter 6 Solutions

The concentrations of all solute species in 0.20 M H 2 S ( a q ) have to be calculated.

Solution Summary: The author explains that the equilibrium constant for the first deprotonation reaction can be written as shown below.

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Interpretation: The concentrations of all solute species in 0.20 M H2S(aq) have to be calculated.

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Chapter 6 Solutions

Interpretation:

The concentrations of all solute species in 0.20 M H2S(aq) have to be calculated.