Concept explainers
A 30.0-mL sample of 0.05 M HClO is titrated by a 0.0250 M KOH solution Ka for HClO is 3.5 × 10−8. Calculate a the pH when no base has been added; b the pH when 30.00 mL of the base has been added; c the pH at the equivalence point; d the pH when an additional 4.00 mL of the KOH solution has been added beyond the equivalence point.
(a)
Interpretation:
The pH of the given points of the titration of
when no base has been added
Concept Introduction:
pOH definition:
The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion
Relationship between pH and pOH:
Answer to Problem 16.135QP
The pH of the given points of the titration of
The pH when no base has been added is 4.4
Explanation of Solution
To Calculate: The pH when no base has been added
Given data:
The volume of
The concentration of
The concentration of
The
pH prior to the start of the titration
Construct an equilibrium table for the hydrolysis of
|
|||
Initial |
0.05
0.05-x |
0.00 | 0.00 |
Change |
|
|
|
Equilibrium |
x | x |
The
Now substitute equilibrium concentrations into the equilibrium-constant expression.
Here, x gives the concentration of hydronium ion
Finally calculate pH as follows,
The pH when no base has been added was calculated as 4.4
(b)
Interpretation:
The pH of the given points of the titration of
when 30.0 mL of the base has been added
Concept Introduction:
pOH definition:
The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion
Relationship between pH and pOH:
Answer to Problem 16.135QP
The pH of the given points of the titration of
The pH when 30.0 mL of the base has been added is 7.5
Explanation of Solution
To Calculate: The pH when 30.0 mL of the base has been added
After the addition of 30.0 mL of 0.0250 M
At this point, the total volume is,
Now, the moles of
After the reaction, the moles of
The moles of
The concentrations are:
Solve for
The pH is calculated as follows,
The pH when 30.0 mL of the base has been added was calculated as 7.5
(c)
Interpretation:
The pH of the given points of the titration of
At the equivalence point
Concept Introduction:
pOH definition:
The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion
Relationship between pH and pOH:
Answer to Problem 16.135QP
The pH of the given points of the titration of
The pH at the equivalence point is 9.8
Explanation of Solution
To Calculate: The pH at the equivalence point
Calculate the volume of
The volume of
Hence, the total volume is as follows,
The equilibrium reaction is,
The value of
Here, x gives the hydroxide ion concentration.
The pH can be calculated from pOH as follows,
The pH at the equivalence point was calculated as 9.8
(d)
Interpretation:
The pH of the given points of the titration of
When an additional 4.00 mL of the
Concept Introduction:
pOH definition:
The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion
Relationship between pH and pOH:
Answer to Problem 16.135QP
The pH of the given points of the titration of
The pH when an additional 4.00 mL of the
Explanation of Solution
To Calculate: The pH when an additional 4.00 mL of the
Calculate the moles of base added.
The total volume after the addition of 4.00 mL of the
The hydroxide ion concentration and the pH are:
The pH is calculated as follows,
The pH when an additional 4.00 mL of the
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Chapter 16 Solutions
General Chemistry - Standalone book (MindTap Course List)
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