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.

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6 0

1 year ago

For each trial, compute the mol of titrant; (molarity x L) and keep the number of significant figures to 4.

MrMuchimi

Answer:

Trial Number of moles

1 0.001249mol

2 0.001232mol

3 0.001187 mol

Explanation:

To calculate the number of moles of tritant you need its molarity.

Since the molarity is not reported, I will use 0.1000M (four significant figures), which is used in other similar problems.

Molarity is the concentration of the solution in number of moles of solute per liter of solution.

In this case the solute is NaOH.

The formula is:

$Molarity=\dfrac{\text{Number of moles of solute}}{\text{Volume of solution in liters}}$

Solve for the number of moles:

$\text{Number of moles}=Molarity\times Volume\text{ }in\text{ }liters$

Then, using the molarity of 0.1000M and the volumes for each trial you can calculate the number of moles of tritant.

Trial mL liters Number of moles

1 12.49 0.01249 0.01249liters × 0.1000M = 0.001249mol

2 12.32 0.01232 0.01232liters × 0.1000M = 0.001232mol

3 11.87 0.01187 0.01187liters × 0.1000M = 0.001187 mol

3 0

3 years ago

Read 2 more answers

PLEASE HELP!! due soon

mina [271]

Answer:

The pic won't load sorry try posting this question again and then i'll be able to answer!

Explanation:

4 0

3 years ago