Sketch the titration curve for the titration of a generic weak base B with a strong acid. The titration reaction is
1 answer:
Answer:
Kindly check the attached picture.
Explanation:
Titration is one of the useful concept or practical in chemistry which is used in the determination of the concentration of a particular chemical specie. Without mincing words, let us dive straight into the solution to the question;
(a) Determining the equivalence point using the graph.
=> The equivalence point is the point at which there has been been "enough" addition of the acid to the base to give us the congugate acid. In the diagram showing the curve, it is the point labelled (a).
(b)The next thing is to determine the region with maximum buffering.
=> The position with the maximum buffering ranges from the point labelled (b) in the graph and it is the point when the acid and the base has been added. But the point with the maximum buffering is labeled as point (c).
(c). Determination of the point where pH = pKb.
=> The Point in which the pH = pKb is the point a
(d). The region where pH depends only on [B].
=> The starting point in the point in which the pH is dependent on the concentration of the base, [B].
(e). The region where pH depends only on [BH+].
=> This is the same point with point labelled as (a). Thus, point (a) and (e) are the same.
(f). The region where pH depends only on the amount of excess strong acid is the point labelled as (f) which is known as the post equivalence point.
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Answer:
The value of the equilibrium constant = 5.213
Explanation:
Here (equilibrium constant) is referred to as the partial pressure of product divided by the partial pressure of reactant with each pressure term raised to power that is equal to its stoichiometric coefficient in balanced equation
.
As such only gas appear in expression as solids takes a value of 1;
SO ; in the given equation from the question:
2 A (g) + B (s) ----> 2 C(s) + D (g)
The value of the equilibrium constant = 5.213
Answer:
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄+ 2H₂O
Explanation:
PbO₂ + MnSO₄ + HNO₃ ⟶ HMnO₄ + PbSO₄ + Pb(NO₃)₂ + H₂O
It will be easiest to balance this equation by the ion-electron method.
1. Write the ionic equation
PbO₂ + Mn²⁺ + SO₄²⁻ + H⁺ + NO₃⁻ ⟶ H⁺ + MnO₄⁻ + Pb²⁺ + SO₄²⁻ + Pb²⁺ + NO₃⁻ + H₂O
2. Eliminate H⁺, H₂O, and spectator ions
PbO₂ + Mn²⁺ ⟶ MnO₄⁻ + Pb²⁺
3. Separate the skeleton equation into two half-reactions.
PbO₂ ⟶ Pb²⁺
Mn²⁺ ⟶ MnO₄⁻
4. Balance all atoms other than H and O
Done
5. Balance O by adding water molecules to the deficient side
PbO₂ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻
6. Balance H by adding H⁺ ions to the deficient side.
PbO₂+ 4H⁺ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺
7. Balance charge by adding electrons to the deficient side.
PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e-
8. Multiply each half-reaction by a number to equalize the electrons transferred.
5 × [PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O]
2 × [Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e⁻]
9. Add the two half-reactions.
5PbO₂+ 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 10H₂O
<u> 2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻ </u>
5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻
10. Cancel species that occur on each side of the equation
5PbO₂ + 2Mn² + <u>8H₂O</u> + <u>20H⁺</u> + <u>10e⁻</u> ⟶ 5Pb²⁺ + 2MnO₄⁻ + <u>10H₂O</u> + <u>16H⁺</u> + <u>10e⁻ </u>
becomes
5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O
11. Add the missing spectator ions
5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O
+ 2SO₄²⁻ + 4NO₃⁻ + 2H⁺ + 2NO₃⁻ +2SO₄²⁻ + 6NO₃⁻ + 2H⁺
becomes
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O
12. Check that all atoms are balanced.
Everything checks. The balanced equation is
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O