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) to precipitate 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

The given equation is :-

Cr³⁺(aq)+3NaOH(aq) → Cr(OH)3(s)+3Na⁺(aq)

From the reaction we can predict that there will be three mole of hydroxide ion will be required for one mole precipitation of chromium.

Volume of chromium ion = 0.60 L

Molarity of chromium ion = 0.013 M

Moles of chromium ion = 0.0078 mol

So, the number of Sodium hydroxide required will be = 3 x 0.0078 =0.0234 moles

Molarity of sodium hydroxide = 1.26 M

Volume of Sodium hydroxide required will be :- 0.0234/ 1.26 = 0.0186 L = 18.6 ml