If 845mL of a 2.5x10^-5 mol/L solution of Ni(NO3)2 was mixed with 195mL of a 4.86x10^-2 mol/L solution of KOH, would precipitate form? Calculate the ion product for the potential precipitate. The Ksp of Ni(OH)2(s) is 6.0x10^-16
845ml(2.5x10^-5M Ni(NO3)2 + 195ml(0.0486M KOH)
=>0.845(2.5x10^-5)mole Ni(NO3)2 + 195(0.0486)mole KOH
=>(2.113x10^-5)mole Ni(NO3)2 + (9.48x10^-3)mole KOH
=>(2.113x10^-5)mole Ni^2 + (9.48x10^-3)mole OH^-
=>(2.113x10^-5 mole/1.04 Liter Soln)Ni^2 + (9.48x10^-3 mole OH^-/1.04 Liter Soln)OH^-
=> (2.03x10^-8 M)Ni^+2 + (9.12x10^-6 M)OH^- at mixing ...
=> Qsp = [Ni^+2][OH^-]^2
------ = (2.03x10^-8)(9.12x10^-6)^2
------ = 2x10^-18
Ksp=6x10^-16 > Qsp=2x10^-18 => No ppt forms
To determine if a precipitate will form when the two solutions are mixed, we need to consider the solubility product constant (Ksp) of Ni(OH)2. If the ion product exceeds the Ksp, then a precipitate will form.
First, let's calculate the number of moles of Ni(NO3)2 and KOH in each solution.
For Ni(NO3)2 solution:
Volume (V1) = 845 mL = 0.845 L
Concentration (C1) = 2.5x10^-5 mol/L
Number of moles (n1) = V1 * C1 = 0.845 L * 2.5x10^-5 mol/L = 2.1125x10^-5 mol
For KOH solution:
Volume (V2) = 195 mL = 0.195 L
Concentration (C2) = 4.86x10^-2 mol/L
Number of moles (n2) = V2 * C2 = 0.195 L * 4.86x10^-2 mol/L = 9.477x10^-3 mol
Next, let's determine the limiting reagent by comparing the number of moles of each reactant. The reactant with the smaller number of moles will be completely consumed, and the other reactant will be in excess.
In this case, Ni(NO3)2 has a smaller number of moles (2.1125x10^-5 mol) compared to KOH (9.477x10^-3 mol). Therefore, Ni(NO3)2 is the limiting reagent.
Now, let's calculate the moles of hydroxide ions (OH-) that can be produced from the reaction between Ni(NO3)2 and KOH.
The balanced chemical equation for the reaction is:
Ni(NO3)2 + 2KOH → Ni(OH)2 + 2KNO3
From the equation, we can see that 1 mole of Ni(NO3)2 will react with 2 moles of KOH to produce 1 mole of Ni(OH)2.
Since the limiting reagent is Ni(NO3)2 and 1 mole of Ni(NO3)2 will react with 2 moles of KOH, the moles of Ni(OH)2 that can be formed is equal to the moles of Ni(NO3)2.
Therefore, the moles of Ni(OH)2 formed = 2.1125x10^-5 mol.
Now, let's calculate the concentration of Ni(OH)2 in the final solution.
The total volume of the final solution is the sum of the volumes of both solutions:
Vtotal = V1 + V2 = 0.845 L + 0.195 L = 1.04 L
The concentration of Ni(OH)2 in the final solution is:
Cfinal = (moles of Ni(OH)2 formed) / Vtotal
= (2.1125x10^-5 mol) / 1.04 L
≈ 2.03x10^-5 mol/L
Finally, let's calculate the ion product (Q) for Ni(OH)2 using the concentration of Ni(OH)2:
Q = [Ni(OH)2]^a
= (Cfinal)^2
= (2.03x10^-5 mol/L)^2
= 4.12x10^-10
Comparing the ion product (Q = 4.12x10^-10) with the solubility product constant (Ksp = 6.0x10^-16), we can see that Q < Ksp.
Since Q < Ksp, no precipitate will form when the solutions are mixed.