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

7th Edition

ISBN: 9781319421175

Author: ATKINS

Publisher: MAC HIGHER

See similar textbooks

Concept explainers

Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electroly…

Videos

Question

Chapter 6, Problem 6E.3AST

Interpretation Introduction

Interpretation:

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 6, Problem 6E.2BST

Chapter 6, Problem 6E.1E

Students have asked these similar questions

Calculate the concentration of HS in an aqueous solution of 0.142 M hydrosulfuric acid, H₂S (aq). [HS] = M.

A 10.0 mL sample of 0.20 mol/L NiF2(aq) is mixed with 20.0 mL of 0.0700 mol/L NaOH(aq) and then diluted to a final volume of 100. mL. Calculate the concentration of OH- ions in the 100 mL mixture before the reaction starts. Express your answer to three significant figures.

Chapter 6 Solutions

CHEMICAL PRINCIPLES (LL) W/ACCESS

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

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

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.

Concept explainers

Videos

Interpretation:

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

Want to see the full answer?

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.

Concept explainers

Videos

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

Want to see the full answer?

Chapter 6 Solutions

Interpretation:

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