
Concept explainers
A 25.0-mL solution of 0.100 M

Interpretation:
The pH values of the solution after the addition of different volumes of
Concept introduction:
Acid–base titration is a technique used to analyze the unknown concentration of the acid or the base through the known concentration of the acid and base.
The equivalence point is the point in the acid–base titration of the chemical reaction where the number of moles of the titrant and the unknown concentration of the analyte are equal. It is used to identify the unknown concentration of the analyte.
The value of pH expresses the acidity or alkalinity of a solution on a logarithmic scale.
When the value of pH is equal to
The formula to calculate the pH of a solution is:
pOH is a measure of the basicity of a solution, which depends on the concentration of hydroxide ions and the temperature of the solution.
The formula to calculate the pOH of a solution is:
The relation between pH and pOH is as:
A base dissociation constant, Kb, (also known base-ionization constant) is a quantitative measure of the strength of an base in solution.
The ionization of a weak base can be represented by the equation:
Where B is weak base and
The base ionization constant for a weak base is:
Where
An acid dissociation constant, Ka, (also known as acidity constant, or acid-ionization constant) is a quantitative measure of the strength of an acid in solution
The dissociation of a weak acid in an aqueous solution:
The acid ionization constant for a weak acid is:
The number of moles of compound initially present in the solution is calculated as:
The molarity of a compound is given by the expression as:
The pH of the buffer solution by the Henderson–Hasselbalch equation can be calculated as:
Answer to Problem 35QP
Solution:
a)
b)
c)
d)
e)
Explanation of Solution
Given information: The concentration of
Explanation:
a)
The neutralization reaction between
From the above equation, 1 mole of
Let
The initial ionization change table for the ionization of
The equilibrium expression for the reaction is represented as follows:
Here,
Substitutes the values of
The value of
Concentration of
The value of
Substitute the value of
Hence, the
Explanation:
b)
The number of moles of
Substitute the values of the concentration and the volume of
The number of moles of
Substitute the values of the concentration and the volume of
On mixing the two solutions, the molarity of the solution changes, but the number of moles will remain the same.
The changes in the number of moles of
Determine the pH of the buffer solution by the Henderson–Hasselbalch equation. This is a basic buffer.
The pH of the buffer is calculated as follows:
Hence, the
Explanation:
c)
The number of moles of
Substitute the values of the concentration and the volume of
The number of moles of
Substitute the values of the concentration and the volume of
On mixing the two solutions, the molarity of the solution changes, but the number of moles will remain the same.
The changes in the number of moles of
Determine the pH of the buffer solution by the Henderson–Hasselbalch equation.
The pH of the buffer is calculated as follows:
Hence, the
Explanation:
d)
The number of moles of
Substitute the values of the concentration and the volume of
The number of moles of
Substitute the values of the concentration and the volume of
On mixing the two solutions, the molarity of the solution changes, but the number of moles will remain the same.
The changes in the number of moles of
At this stage, an equivalence point of the titration is reached.
The molarity of
Substitute the values of the number of moles and the volume of
Let
The initial concentration change for the ionization of
The equilibrium expression for the reaction is written as follows:
Here,
Substitute the values of
The value of x is very small as compared to
Concentration of
The
Substitute the value of
The value of
Hence, the
e)
Given information: The concentration of
Explanation:
The number of moles of
Substitute the values of the concentration and the volume of
The number of moles of
Substitute the values of the concentration and the volume of
On mixing the two solutions, the molarity of the solution changes, but the number of moles will remain the same.
The changes in the number of moles of
The molarity of
Substitute the values of the number of moles and the volume
The concentration of
The
Substitute the value of
The value of
Substitute the value of
Hence, the
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Chapter 17 Solutions
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