Explanation H N O 2 is a weak acid and its conjugate base is N O 2 − . The required equilibrium is given below. H N O 2 ( a q ) + H 2 O ( l ) ⇌ H 3 O + ( a q ) + N O 2 − ( a q ) K a = [ H 3 O + ] [ N O 2 − ] [ H N O 2 ] = 4.3 × 10 − 4 Given that, the buffer contains 0.15 M H N O 2 ( a q ) and 0.20 M N a N O 2 ( a q ) . The equilibrium concentration of H 3 O + ions can be calculated as shown below. K a = [ H 3 O + ] [ N O 2 − ] [ H N O 2 ] = 4

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Chapter 6, Problem 6G.1AST

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The pH of a buffer solution that is 0.15 M HNO2(aq) and 0.20 M NaNO2(aq) has to be calculated.

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Chapter 6, Problem 6F.10E

Chapter 6, Problem 6G.1BST

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Chapter 6 Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

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 p H of a buffer solution that is 0.15 M H N O 2 ( a q ) and 0.20 M N a N O 2 ( a q ) has to be calculated.

Solution Summary: The author explains that the pH of a buffer solution that contains 0.15MTextHNO_2(aq) and 020

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Interpretation:

The pH of a buffer solution that is 0.15 M HNO2(aq) and 0.20 M NaNO2(aq) has to be calculated.

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Chapter 6 Solutions

The p H of a buffer solution that is 0.15 M H N O 2 ( a q ) and 0.20 M N a N O 2 ( a q ) has to be calculated.

Solution Summary: The author explains that the pH of a buffer solution that contains 0.15MTextHNO_2(aq) and 020

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Interpretation: The pH of a buffer solution that is 0.15 M HNO2(aq) and 0.20 M NaNO2(aq) has to be calculated.

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Chapter 6 Solutions

Interpretation:

The pH of a buffer solution that is 0.15 M HNO2(aq) and 0.20 M NaNO2(aq) has to be calculated.