A solution of .12 L of 0.160 M KOH is mixed with a solution of .3 L of 0.230 M NiSO4.

the equation for this reaction is:
2KOH (aq)+NiSO4 (aq) ----->K2SO4 (aq)+Ni(OH)2 (s)

.89 grams of Ni(OH)2 precipitate form

I need to know the concentration remaining in the solution of:
1) Ni (II)
2) SO4
3) K

I do not know how to do these nor can i find aid to show me how to do it step by step, so to whoever answers this question, could you also show steps so i can learn how to do this? thank you to whoever puts in their time and helps me!!

See your other post above.

Sure! I'd be happy to help you step by step.

To find the concentration remaining in the solution of Ni(II), SO4, and K, we need to use the stoichiometry of the balanced chemical equation and the principle of molar ratios.

Step 1: Determine the moles of KOH and NiSO4 in the respective solutions.
Moles of KOH = volume (L) x concentration (M) = 0.12 L x 0.160 M = 0.0192 moles
Moles of NiSO4 = volume (L) x concentration (M) = 0.3 L x 0.230 M = 0.069 moles

Step 2: Determine the limiting reagent.
The limiting reagent is the one that will be completely consumed based on its stoichiometric ratio with the other reactant. In this case, we have a 2:1 ratio between KOH and NiSO4. So, for every 2 moles of KOH, we need 1 mole of NiSO4.
Since we have 0.0192 moles of KOH, we will need 0.0096 moles of NiSO4 to react completely (0.0192/2 = 0.0096).
However, we only have 0.069 moles of NiSO4, which is in excess and will not be completely consumed. Therefore, NiSO4 is the excess reagent, and KOH is the limiting reagent.

Step 3: Determine the moles of the other reactant consumed based on the limiting reagent.
Since the stoichiometric ratio between KOH and NiSO4 is 2:1, this means for every 2 moles of KOH consumed, 1 mole of NiSO4 will be consumed.
Therefore, moles of NiSO4 consumed = 0.0192 moles KOH / 2 = 0.0096 moles.

Step 4: Determine the moles of product formed based on the limiting reagent.
From the balanced chemical equation, we know that 2 moles of KOH react to form 1 mole of Ni(OH)2 precipitate.
Therefore, moles of Ni(OH)2 formed = 0.0192 moles of KOH / 2 = 0.0096 moles.

Step 5: Calculate the molar mass of Ni(OH)2.
Ni has a molar mass of 58.69 g/mol, and OH (from KOH) has a molar mass of 17.01 g/mol.
So, molar mass of Ni(OH)2 = (58.69 g/mol x 1) + (17.01 g/mol x 2) = 92.71 g/mol.

Step 6: Calculate the mass of precipitate formed.
Mass of Ni(OH)2 = moles of Ni(OH)2 x molar mass of Ni(OH)2 = 0.0096 moles x 92.71 g/mol = 0.883 g.

Step 7: Calculate the concentration remaining in the solution.
Since KOH is the limiting reagent, all of it has reacted. Thus, the concentration of KOH remaining is 0 M.

Let's now proceed to find the concentration of Ni(II), SO4, and K.

1) Concentration of Ni(II):
Moles of Ni(II) = moles of Ni(OH)2 formed (Step 4) = 0.0096 moles
Volume of solution remaining = Volume of NiSO4 solution - Volume of Ni(OH)2 formed = 0.3 L - 0.12 L = 0.18 L
Concentration of Ni(II) = moles of Ni(II) / volume of solution remaining = 0.0096 moles / 0.18 L = 0.053 M

2) Concentration of SO4:
Since NiSO4 completely dissociates into Ni(II) and SO4 ions, the moles of SO4 in the initial solution and the remaining solution will be the same.
Moles of SO4 = moles of NiSO4 in the initial solution = 0.069 moles
Volume of solution remaining = Volume of NiSO4 solution - Volume of Ni(OH)2 formed = 0.3 L - 0.12 L = 0.18 L
Concentration of SO4 = moles of SO4 / volume of solution remaining = 0.069 moles / 0.18 L = 0.383 M

3) Concentration of K:
All of the KOH has reacted, so the concentration of K remaining is 0 M.

Therefore, the concentrations remaining in the solution are:
1) Ni(II): 0.053 M
2) SO4: 0.383 M
3) K: 0 M

I hope this step-by-step explanation helps you understand how to approach and solve similar problems in the future! Let me know if you have any further questions.