Problem 87 - Chemistry: The Central Science

Question

A solution of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH}\) is mixed with a solution of \(200.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{NiSO}_{4}\). (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

Step-by-Step Solution

Answer

The balanced chemical equation for the reaction between KOH and NiSO4 is: \(2 \mathrm{KOH} + \mathrm{NiSO}_{4} \rightarrow \mathrm{K}_{2}\mathrm{SO}_{4} + \mathrm{Ni(OH)}_{2}\) Ni(OH)2 forms a precipitate, KOH is the limiting reactant, and 0.9271 g of Ni(OH)2 precipitate form. The remaining concentrations of ions in the solution are [K+] = 0.0667 M, [Ni2+] = 0.0667 M, and [SO42-] = 0.0333 M.

Step 1: (a) Write the balanced chemical equation for the reaction that occurs.

When KOH and NiSO4 are mixed, a double displacement reaction occurs, leading to the formation of K2SO4 and Ni(OH)2. The balanced chemical equation for the reaction is given by: \(2 \mathrm{KOH} + \mathrm{NiSO}_{4} \rightarrow \mathrm{K}_{2}\mathrm{SO}_{4} + \mathrm{Ni(OH)}_{2}\)

Step 2: (b) Identify the precipitate that forms.

Ni(OH)2 forms a precipitate when KOH and NiSO4 react because of its low solubility in water.

Step 3: (c) Determine the limiting reactant.

First, find the moles of each reactant. Moles of KOH: Moles = Molarity × Volume (in liters) Moles of KOH = 0.200 M × (100.0 mL/1000 mL) = 0.0200 mol Moles of NiSO4: Moles = Molarity × Volume (in liters) Moles of NiSO4 = 0.150 M × (200.0 mL/1000 mL) = 0.0300 mol Next, determine the mole ratio between KOH and NiSO4 using the stoichiometric coefficients from the balanced chemical equation: Mole ratio = 2 (KOH) : 1 (NiSO4) Compare the actual mole ratio with the stoichiometry: Actual ratio = moles of KOH / moles of NiSO4 = 0.0200 / 0.0300 = 0.6667 Since the actual mole ratio is less than the stoichiometric requirement, KOH is the limiting reactant.

Step 4: (d) Calculate the mass of the precipitate formed.

Since KOH is the limiting reactant, the amount of precipitate formed depends on the amount of KOH: Moles of Ni(OH)2 formed = 1/2 x Moles of KOH = 1/2 x 0.0200 mol = 0.0100 mol Now, we can calculate the mass of Ni(OH)2: Mass = Moles × Molar mass (in g/mol) Molar mass of Ni(OH)2 = 58.69 (Ni) + 2 × (16.00 + 1.008) (OH) = 92.708 g/mol Mass of Ni(OH)2 = 0.0100 mol × 92.708 g/mol = 0.9271 g

Step 5: (e) Calculate the concentration of each ion that remains in solution.

First, determine the moles of each product after the reaction. Moles of K2SO4 formed = Moles of KOH / 2 = 0.0200 mol / 2 = 0.0100 mol Moles of remaining NiSO4 = Initial moles of NiSO4 - moles of KOH / 2 = 0.0300 mol - 0.0100 mol = 0.0200 mol Moles of K+ ion = 2 × moles of K2SO4 formed = 2 × 0.0100 mol = 0.0200 mol Total volume of the solution = 100 mL (KOH) + 200 mL (NiSO4) = 300 mL Now, calculate the molarities of the remaining ions: Molarity of K+ ion: 0.0200 mol / (300 mL / 1000 mL) = 0.0667 M Molarity of Ni2+ ion: 0.0200 mol / (300 mL / 1000 mL) = 0.0667 M Molarity of SO42- ion: 0.0100 mol / (300 mL / 1000 mL) = 0.0333 M Thus, the remaining concentrations of ions in the solution are [K+] = 0.0667 M, [Ni2+] = 0.0667 M, and [SO42-] = 0.0333 M.