J.R. S. answered • 03/12/17
Tutor
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Ph.D. University Professor with 10+ years Tutoring Experience
a) 2KOH + NiSO4 ==> Ni(OH)2 + K2SO4
b) moles KOH = 0.1 L x 0.200 mol/L = 0.0200 moles KOH
moles NiSO4 = 0.2 L x 0.15 mol/L = 0.03 moles NiSO4
limiting reactant is KOH since it takes 2x the amount of moles as there are in NiSO4, or 0.06 needed, but have only 0.02.
c) 2 mol KOH ==> 1 mole Ni(OH)2 which is the solid product
0.0200 moles KOH x 1 mole Ni(OH)2/2 moles KOH = 0.010 moles Ni(OH)2 formed
0.01 moles Ni(OH)2 x 92.7 g/mole = 0.927 g Ni(OH)2 formed
d) For calculating final concentrations, recall that the final volume is now 300 ml (0.3 L)
[SO4] = 0.03 moles/0.03 L = 0.1 M
[K+] = 2x0.033M = 0.067 M K^+
Ni^2+ = 0.067 M
J.R. S.
tutor
That would be absolutely correct. My oversight and carelessness. Thanks. I'll fix it for future reference.
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09/17/18
Candace H.
No worries! Thank you--:)
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09/17/18
Bill R.
Why is Hydroxide not listed in the final concentration count?
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09/24/19
J.R. S.
tutor
The hydroxide ion is all tied up in the precipitate as Ni(OH)2 so isn’t included in the soluble ions.
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09/24/19
Nathan F.
Could you explain how you calculate the concentration of Nickel? Also, why is that considered part of the solution if it is part of the precipitate?
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11/18/19
J.R. S.
tutor
You calculate the final conc. of Ni ion by finding how much is left over AFTER the reaction. Not all of the Ni is precipitated because the NiSO4 is in excess. Initial moles of Ni = 0.03 (from NiSO4). Moles of Ni reacted = 0.01 (reacts with KOH in 1:2 ratio). Moles Ni remaining = 0.03 - 0.01 = 0.02 moles. Final volume = 0.3 L. Final [Ni ion] = 0.02mol/0.3L = 0.067M
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11/18/19
Candace H.
SO4 = 0.1 M (all 0.03 moles would be in solution at the end 0.03moles/.3L = .1M)
09/17/18