Answer 376.4ml Explanation First determine the no of mole of Ni2+ present in the

Question

Answer

376.4ml

Explanation

First determine the no of mole of Ni2+ present in the NiI2

solution

Concentration of NiI2 = 0.789M = 0.789 mole / L

Volume of NiI2 = 156ml

No of mole NiI2 = (0.789mol/1000ml)×156ml =0.12308mol

NiI2 molecule having 1Ni2+

No of mole of Ni2+ = 0.12308

Now consider the reaction

NiI2(aq) + 2NaOH(aq) --------> Ni(OH)2(s) + 2NaI

to precipitate 1mole of Ni2+ , 2mole of OH- required

So, 0.12308mole of Ni2+ require 0.24616mole of OH-

Concentration of NaOH = 0.654mole/L

NaOH is strong base , so

Concentration of NaOH = Concentration of OH-

Therefore,

Volume of NaOH required = (1000ml/0.654mol)×0.24616mol = 376.39ml

Explanation / Answer

er of milliliters of 0.654 M NaOH required to precipitate ll fthe Ni2* ions in 156 ml of 0.789 M Nil solution as NiKOH), The equation for the reaction is: Nil(a)+2NaOH(a)N(OH)(s)+2Nal(ag) mL NaOH Submit Answer Retry Entire Group 9 more group attempts remaining

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