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madreJ [45]
1 year ago
6

Precipitation calculations with Ni2+ and Pb2+

Chemistry
1 answer:
Maurinko [17]1 year ago
6 0

Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.

When Ni(OH)₂ starts precipitate :

Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]

5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]

[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18

[ OH⁻ ] = 5.5 × 10⁻⁸ M

pOH = 7.2

therefore , pH = 14 - 7.2

                  pH = 6.8

Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.

To learn more about pH here

brainly.com/question/15289741

#SPJ1

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Explanation:

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If it takes 38.70cm of 1.90M NaOH to neutralize 10.30cm of H2SO4 in a battery, what is the molarity of H2SO4?
slamgirl [31]

Answer:

The molarity of the acid, H₂SO₄ is 3.57 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

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Molarity of acid, H₂SO₄ (Mₐ) =?

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Cross multiply

Mₐ × 10.3 × 2 = 73.53 × 1

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Divide both side by 20.6

Mₐ = 73.53 / 20.6

Mₐ = 3.57 M

Thus, the molarity of the acid, H₂SO₄ is 3.57 M

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