Explanation H 2 S O 4 is a polyprotic acid. It is strong acid in its first deprotonation. Therefore, to estimate its p H , the second deprotonation needs to be taken in account. The first and second deprotonation reactions are given below. H 2 S O 4 ( a q ) + H 2 O ( l ) → H 3 O + ( a q ) + H S O 4 − ( a q ) H S O 4 − ( a q ) + H 2 O ( l ) ⇌ H 3 O + ( a q ) + S O 4 2 − ( a q ) , K a 2 = 0.012 The expression for equilibrium constant for the above reaction can be written as shown below, K a 2 = [ H 3 O + ] [ S O 4 2 − ] [ H S O 4 − ] The conjugate base H S O 4 − is amphiprotic and it contributes protons to the solution which has been shown in the second deprotonation reaction above. An equilibrium table can be set up as given below. C o m p o s i t i o n ( M ) H S O 4 − H 2 O H 3 O + S O 4 2 − I n i t i a l c o n c e n t r a t i o n 0.10 − 0.10 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

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Chapter 6, Problem 6E.1BST

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The pH of 0.10 M H2SO4(aq) has to be calculated.

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Chapter 6, Problem 6E.1AST

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The p H o f 0.10 M H 2 S O 4 ( a q ) has to be calculated.

Solution Summary: The author explains the calculation of the pH of a polyprotic acid, which is strong in its first deprotonation. The conjugate base HSO_4- is amphi

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Interpretation:

The pH of 0.10 M H2SO4(aq) has to be calculated.

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Chapter 6 Solutions

The p H o f 0.10 M H 2 S O 4 ( a q ) has to be calculated.

Solution Summary: The author explains the calculation of the pH of a polyprotic acid, which is strong in its first deprotonation. The conjugate base HSO_4- is amphi

Concept explainers

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Interpretation: The pH of 0.10 M H2SO4(aq) has to be calculated.

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Chapter 6 Solutions

Interpretation:

The pH of 0.10 M H2SO4(aq) has to be calculated.