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Amy H.

asked • 09/25/16

Chemistry Problem

A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4.
(a) Write the balanced chemical equation for the reaction that occurs.
(b) Identify the limiting reactant?
(c) How many grams of solid product form?
(d) What is the concentration of each ION that remain after the reaction?

1 Expert Answer

By:

Candace H.

If KOH is the limiting reactant, that means there would be Nickel ions that are not part of the precipitate.  There would also be a higher concentration of sulfate because, that too, is in excess.
Ni = 0.067 M
SO4 = 0.1 M (all 0.03 moles would be in solution at the end 0.03moles/.3L = .1M)
K+= 0.067M
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09/17/18

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

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