Question

Suppose 1.87g of nickel(II) bromide is dissolved in 200.mL of a 52.0mM aqueous solution of potassium carbonate. Calculate the final molarity of nickel(II) cation in the solution. You can assume the volume of the solution doesn't change when the nickel(II) bromide is dissolved in it. Be sure your answer has the correct number of significant digits. The answer is needed in M. Thank you.

Answer 1

Ans :-

Initial number of Moles of NiBr₂= 1.87g / 118.702 g/mol = 0.01576 mol

Initial number of moles of K₂CO₃ = Molarity x Volume = 52.0 mM x 200.0 mL = 0.052 mo/L x 0.200 L = 0.0104 mol

ICF table is :

..........................NiBr₂ (aq)..............+.............K₂CO₃ (aq) ------------> NiCO₃ (s)............+..................2KBr (aq)

Initial (I)..............0.01576 mol.........................0.0104 mol......................0.0 mol....................................0.0 mol

Change (C)..........-0.0104 ................................-0.0104.........................+0.0104...................................+0.0104

Final (F)...............0.00536 mol..........................0.0 mol.........................0.0104 mol...............................0.0104 mol

Because NiCO₃ is sparingly soluble salt (K_sp of NiCO₃ = 6.6 x 10^-9 ), therefore concentration of Ni²⁺ in NiCO₃ is very very less. So, Number of moles of Ni²⁺ in NiCO₃ can be consider as negligible.

Now,

Number of moles of Ni²⁺ cation in NiBr₂ = 0.00536 mol

Volume = 200.0 mL

[Ni²⁺] = 0.00536 mol/0.200 L = 0.268 M