Explanation The addition of N a O C l to water results into a basic solution. The reaction is given below. O C l − ( a q ) + H 2 O ( l ) ⇌ H O C l ( a q ) + O H − ( a q ) The base ionization constant for the above equation can be written as given below. K b = [ H O C l ] [ O H − ] [ O C l − ] The concentration of H 3 O + can be calculated from the p H as given below. p H = − log [ H 3 O + ] ⇒ [ H 3 O + ] = 10 − p H = 10 − 5.50 = 3.2 × 10 − 6 . A table can be set up as given below. O C l − ( a q ) H O C l ( a q ) O H − ( a q ) I n i t i a l c o n c . ( M ) 0.010 0 1.0 × 10 − 7 C h a n g e i n c o n c . ( M ) − x + x + x E q u i l i b r i u m c o n c

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ISBN: 9781285199047

Author: John W. Moore, Conrad L. Stanitski

Publisher: Cengage Learning

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Chapter 14, Problem 100QRT

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Interpretation:

The pH of a 0.010 M solution of NaOCl(Kb=1.5×10−7) has to be calculated.

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Chapter 14, Problem 99QRT

Chapter 14, Problem 101QRT

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The p H of a 0.010 M solution of N a O C l ( K b = 1.5 × 10 − 7 ) has to be calculated.

Solution Summary: The author calculates the pH of a 0.010M solution of NaOClleft.

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Interpretation:

The pH of a 0.010 M solution of NaOCl(Kb=1.5×10−7) has to be calculated.

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Chapter 14 Solutions

The p H of a 0.010 M solution of N a O C l ( K b = 1.5 × 10 − 7 ) has to be calculated.

Solution Summary: The author calculates the pH of a 0.010M solution of NaOClleft.

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Interpretation: The pH of a 0.010 M solution of NaOCl(Kb=1.5×10−7) has to be calculated.

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Chapter 14 Solutions

Interpretation:

The pH of a 0.010 M solution of NaOCl(Kb=1.5×10−7) has to be calculated.