
Interpretation:
The equivalence volume of
Concept introduction:
Volumetric principle:
The volume and concentration of unknown solution is determined by it is titrate with known volume and concentration solution.
The volume and concentration of unknown solution is required equivalent volume and concentration of known solution in the volumetric titration.
Where,
Degree of dissociation:
The ratio of mole of reactant that underwent to dissociating to mole of initial reactant is known as degree of dissociation.
In EDTA the degree of dissociation is,
If pH is fixed, the degree of dissociation
Where,

Answer to Problem 12.7P
The volume of solution at equivalent point is
The conditional formation value is
Explanation of Solution
To determine the volume of solution at equivalent point in
Given,
Volume of known solution is
Concentration of known solution is
Concentration of
According to the volumetric principle,
The given values are plugged in above equation to give a volume of solution at equivalent point in
To give the degree of dissociation
Given,
From the standard data table the
The conditional formation value is
The equivalence volume of
(a)
Interpretation:
The
Conditional Formation Constant:
In the reaction of metal with ligand, the equilibrium constant is called as formation constant or the stability constant.
The formation constant for above complex reaction is,
Where,
If pH is constant, the degree of dissociation
Where,
(a)

Answer to Problem 12.7P
The
Explanation of Solution
To determine the
Given,
Equivalent point in
Volume of different is
The volume difference and concentration are plugged in above equation to give
The
(b)
Interpretation:
At the equivalent point, the
Concept Information:
Degree of dissociation:
The ratio of mole of reactant that underwent to dissociating to mole of initial reactant is known as degree of dissociation.
In EDTA the degree of dissociation is,
If pH is fixed, the degree of dissociation
Where,
(b)

Explanation of Solution
To calculate the
The equilibrium reaction is,
The concentration of
From the above calculation the
Hence, the
The
The
At the equivalent point, the
(c)
Interpretation:
At
Concept Information:
Conditional Formation Constant:
In the reaction of metal with ligand, the equilibrium constant is called as formation constant or the stability constant.
The formation constant for above complex reaction is,
Where,
If pH is constant, the degree of dissociation
Where,
(c)

Answer to Problem 12.7P
At
Explanation of Solution
To calculate the
Given,
To calculate the
The equilibrium reaction is,
The concentration of
Formal concentration of EDTA is,
The
The calculated EDTA concentration and
At
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Chapter 12 Solutions
Quantitative Chemical Analysis 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card)
- -AG|F=2E|V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: Acidic solution -0.93 +0.38 -0.51 -0.06 H3PO4 →H4P206 H3PO3 H3PO2 → P→ PH3 -0.28 -0.50 → -0.50 Basic solution 3-1.12 -1.57 -2.05 -0.89 PO HPO →→H2PO2 P PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P2O6 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH, 0.0 -0.5- 2 3 9 3 -1.5 -2.0 Pa H,PO H,PO H,PO -3 -1 0 2 4 Oxidation state, N 2 b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) c) Elemental phosphorus tends to disproportionate under basic conditions. Use data in…arrow_forwardThese two reactions appear to start with the same starting materials but result in different products. How do the chemicals know which product to form? Are both products formed, or is there some information missing that will direct them a particular way?arrow_forwardWhat would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 3 1 2 2. n-BuLi • Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure. 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Priva ×arrow_forward
- Predict the products of this organic reaction: Explanation Check IN NaBH3CN H+ ? Click and drag to start drawing a structure. D 5 C +arrow_forwardPredict the products of this organic reaction: H3O+ + ? • Draw all the reasonable products in the drawing area below. If there are no products, because no reaction will occur, check the box under the drawing area. • Include both major and minor products, if some of the products will be more common than others. • Be sure to use wedge and dash bonds if you need to distinguish between enantiomers. No reaction. Click and drag to start drawing a structure. dmarrow_forwardIarrow_forward
- Draw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option commandarrow_forwardAG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, Narrow_forward4. For the following complexes, draw the structures and give a d-electron count of the metal: a) Tris(acetylacetonato)iron(III) b) Hexabromoplatinate(2-) c) Potassium diamminetetrabromocobaltate(III) (6 points)arrow_forward
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