Let us assume that Cr(OH)3(s) is completely insoluble, which signifies that the

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. If you had a 0.600 L solution containing 0.0130 M of Cr3+(aq), and you wished to add enough 1.29 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.

Explanation / Answer

V = 0.6 L solution of M = 0.013 Cr+3

M = 1.29 M of NaOH

min amount of NaOH must be added

Cr3+ + 3NAOH --> Cr(OH)3 + 3Na+

find molees of Cr+3

mol = M*W = 0.013*0.6 = 0.0078 mol of Cr+3

therefore

we need 3X that of NAOh

3*0.0078 = 0.0234 mol of NaOH

M = mol/V

V = mol/M = (0.0234 ) /1.29 = 0.018139 L or 18.139 ml

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