Question

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.

Answer 1

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

Step-by-step explanation:

`Cr^(3+)(aq)+3NaOH(aq) \rightarrow Cr(OH)_3(s)+3Na^+(aq)`

Molarity of chromium ions ,M= 0.0130 M

Volume of chromium ions ,V= 0.600 L

Moles of chromium ions = n

`Molarity=(Moles)/(Volume (L))`

`M=(n)/(V)`

`0.0130 M=(n)/(0.600 L)=0.0078 mol`

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

Then 0.0078 moles of chromium ions will react with :

`(3)/(1)* 0.0078 mol=0.0234 mol` of NaOH

Moles of NaOH = n' = 0.0234 moles

Molarity of NaOH solution = M'= 1.26 M

Volume of the NaOH solution = V' = ?

`M'=(n')/(V')`

`1.26 M=(0.0234 mol)/(V')`

`V'=1.26 M* 0.0234 mol= 0.029484 L \approx 0.0295 L`

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