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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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The sum of -(1) + (2) + (3) gives:

<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

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<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>

<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>

<em />

I hope it helps!

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