Concept explainers
Interpretation:
The pH for the titration of
Explanation of Solution
Given,
Titration of
Addition of
Given system is titration of
The concentration of
Addition of
Given system is titration of
The given titration is shown below,
Moles of
Moles of
Limiting reagent:
According to the balanced chemical equation, for 1 mole of
But there is excess amount of
ICE table:
S (mol) | 0 | Excess | ||
R (mol) | ||||
F(mol) | 0 | Excess |
The total volume of the given solution is 1 ml + 0.50 ml = 1.50 ml (or)
One mole of
The concentration of
Addition of
Given system is titration of
The given titration is shown below,
Moles of
Moles of
Limiting reagent:
According to the balanced chemical equation, for 1 mole of
But there is excess amount of
ICE table:
S (mol) | 0 | Excess | ||
R (mol) | ||||
F(mol) | 0 | Excess |
The total volume of the given solution is 1 ml + 1 ml = 2 ml (or)
One mole of
The concentration of
Addition of
Given system is titration of
The given titration is shown below,
Moles of
Moles of
ICE table:
S (mol) | 0 | Excess | ||
R (mol) | ||||
F(mol) | 0 | 0 | Excess |
The total volume of the given solution is 1 ml + 2.4 ml = 3.4 ml (or)
The resulting solution is neutral. Therefore,
Hence,
The concentration of
Addition of
Given system is titration of
The given titration is shown below,
Moles of
Moles of
Limiting reagent:
According to the balanced chemical equation, for 1 mole of
But there is less amount of
ICE table:
S (mol) | 0 | Excess | ||
R (mol) | ||||
F(mol) | 0 | Excess |
The total volume of the given solution is 1 ml + 3 ml = 4 ml (or)
The concentration of
The titration curve is shown below,
Figure 1
The curve in Question 16.21 has the same general form, but the equivalence point occurs at half the volume the equivalence point occurs in Question 16.23.
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Chapter 16 Solutions
Chemistry: Principles and Practice
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- Consider all acid-base indicators discussed in this chapter. Which of these indicators would be suitable for the titration of each of these? (a) NaOH with HClO4 (b) acetic acid with KOH (c) NH3 solution with HBr (d) KOH with HNO3 Explain your choices.arrow_forwardRepeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M propanoic acid (HC3H5O2,Ka = 1.3 105) with 0.100 M NaOH.arrow_forwardSketch a titration curve for the titration of potassium hydroxide with HCl, both 0.100 M. Identify three regions in which a particular chemical species or system dominates the acid-base equilibria.arrow_forward
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