Net Ionic Equation

Question

Predict the mass of precipitate (Ni(OH)_2(aq)) expected when 50.0 mL of 0.45 M nickel(II) sulfate solution (NiSO_4(aq)) is combined with 25.0 mL of 1.00 M sodium hydroxide solution (NaOH_(aq)).

Answer 1

NiSO_4(aq) + NaOH_(aq) --> Na₂SO_4(aq) + Ni(OH)₂

To balance the equation, multiply 2 in the sodium in the reactants

NiSO_4(aq) + 2NaOH_(aq) --> Na₂SO_4(aq) + Ni(OH)₂

To determine the moles of product produced we must first determine the number of moles of each reactant

Moles of NiSO_4(aq) = (0.0500)(0.45) = 0.0225

Moles of NaOH_(aq) = (0.0250)(1.0) = 0.025

Since 0.0225 moles of NiSO_4(aq) needs to react with 2 times 0.0225= 0.0450 moles of NaOH_(aq), NaOH is the limiting reagent. Thus. 0.250 moles of NaOH react with (0.5)(0.0250) = 0.0125 moles of NiSO₄ to produce 0.0125 moles of Ni(OH)_2.

Thus the mass of precipitate = moles of precipitate times molar mass of Ni(OH)₂ = (0.0125) (58.7+2(16+1)) = (0.0125)(58.7+34) = (0.0125)(92.7) = 1.16 grams of precipitate to three significant figures.

Answer 2

The first step in this problem is to determine the limiting reagent. To do that we need to know how many moles of each reactant we have. To find the amount of moles of each reactant we need to first convert the volume to Liters by dividing by 1000, and then multiplying the molarity (mol/L) by the volume.

NiSO4: .45 M * .05 L = ,0225 mol

NaOH: 1 M * .025 L - .025 mol

It may seem that Ni(II) is the limiting reagent, but remember the problem gave us the formula for the product (Ni(OH)2. That means we need to divide the OH concentration in half to find the effective concentration since the product takes two hydroxide anions. That means only .0125 mol of Ni(OH)2 can be formed.

Now that we know how many moles of product is formed we can easily find the mass of the product. We simply need to multiply by the molecular mass of the compound which is 92.707 g/mol.

.0125 * 92.707 = 1.2 g of Ni(OH)2.

I have not been following the rules of significant figures until the end. The molarity of NiSO4 only has 2 sig figs so I limited my final answer to 2 sig figs.