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Olegator [25]
3 years ago
11

Name each of the following complex ions and identify the oxidation number of the metal: (a) [Fe(CN)6]42; (b) [Co(NH3)6]31; (c) [

Co(CN)5(OH2)]22; (d) [Co(NH3)5(SO4)]1.
Chemistry
1 answer:
xenn [34]3 years ago
5 0

Answer:

(a) [Fe(CN)₆]⁴⁻ - Hexacyanoferrate(II) ion

Central metal atom -  Fe

Oxidation state of central metal Fe = +2

(b) [Co(NH₃)₆]³⁺ : Hexamminecobalt(III) ion

Central metal atom -  Co

Oxidation state of central metal Co = +3

(c) [Co(CN)₅(OH₂)]²⁻ : Aquapentacyanidocobalt(III) ion

Central metal atom -  Co

Oxidation state of central metal Co = +3

(d) [Co(NH₃)₅(SO₄)]⁺ :  Pentaamminesulfatocobalt(III) ion

Central metal atom -  Co

Oxidation state of central metal Co = +3

Explanation:

Oxidation number, also known as oxidation state, describes the number of electrons lost or gained by a chemical element.

(a) [Fe(CN)₆]⁴⁻ - Hexacyanoferrate(II) ion

Central metal atom -  Fe

Let the oxidation state of Fe = x

Charge on CN⁻ ligand = (-1)

x + (-1) × 6 = -4

⇒ x = -4 + 6 = +2

<u>Oxidation state of Fe: x = +2</u>

(b) [Co(NH₃)₆]³⁺ : Hexamminecobalt(III) ion

Central metal atom -  Co

Let the oxidation state of Co = x

Charge on NH₃ ligand = 0

x + (0) × 6 = +3

⇒ x = +3

<u>Oxidation state of Co: x = +3</u>

(c) [Co(CN)₅(OH₂)]²⁻ : Aquapentacyanidocobalt(III) ion

Central metal atom -  Co

Let the oxidation state of Co = x

Charge on CN⁻ ligand = -1

Charge on H₂O ligand = 0

x + (-1) × 5 + 0 × 1 = -2

⇒ x = +3

<u>Oxidation state of Co: x = +3</u>

(d) [Co(NH₃)₅(SO₄)]⁺ :  Pentaamminesulfatocobalt(III) ion

Central metal atom -  Co

Let the oxidation state of Co = x

Charge on NH₃ ligand = 0

Charge on SO₄²⁻ ligand = -2

x + (0) × 5 + (-2) × 1 = +1

⇒ x = +3

<u>Oxidation state of Co: x = +3</u>

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a. Calculate the pH of a solution that has [H3O+]= 4.3 x 10-5 M. b. Is the solution acidic or basic? How do you know?
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* pH=4.37

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Explanation:

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3 years ago
An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
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<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

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[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

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Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

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