Let us assume that Cr(OH)3(s) is completely insoluble, which signifies that the precipitation reaction with NaOH(aq) (presented in the transition) would go to completion.
Cr3+(aq)+3NaOH(aq) → Cr(OH)3(s)+3Na+(aq)
If you had a 0.600 L solution containing 0.0130 M of Cr3+(aq), and you wished to add enough 1.26 M NaOH(aq) toprecipitate all of the metal, what is the minimum amount of the NaOH(aq) solution you would need to add? Assume that the NaOH(aq) solution is the only source of OH−(aq) for the precipitation.
Express the volume to three significant figures and include the appropriate units.

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Tringa0 Tringa0 · Answer 1 · 2019-11-07T09:13:33+01:00

Answer:

0.0295 liters is the minimum amount of the NaOH(aq) solution we would add.

Explanation:

[tex]Cr^{3+}(aq)+3NaOH(aq) \rightarrow Cr(OH)_3(s)+3Na^+(aq)[/tex]

Molarity of chromium ions ,M= 0.0130 M

Volume of chromium ions ,V= 0.600 L

Moles of chromium ions = n

[tex]Molarity=\frac{Moles}{Volume (L)}[/tex]

[tex]M=\frac{n}{V}[/tex]

[tex]0.0130 M=\frac{n}{0.600 L}=0.0078 mol[/tex]

According to reaction , 1 moles of chromium ions reacts with 3 moles of NaOH.

Then 0.0078 moles of chromium ions will react with :

[tex]\frac{3}{1}\times 0.0078 mol=0.0234 mol[/tex] of NaOH

Moles of NaOH = n' = 0.0234 moles

Molarity of NaOH solution = M'= 1.26 M

Volume of the NaOH solution = V' = ?

[tex]M'=\frac{n'}{V'}[/tex]

[tex]1.26 M=\frac{0.0234 mol}{V'}[/tex]

[tex]V'=1.26 M\times 0.0234 mol= 0.029484 L \approx 0.0295 L[/tex]

0.0295 liters is the minimum amount of the NaOH(aq) solution we would add.