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?

Short answer

(a) The balanced chemical equation for the reaction is: $2\mathrm{KOH}+{\mathrm{NiSO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+\mathrm{Ni}(\mathrm{OH}{)}_2$ (b) The precipitate formed is nickel(II) hydroxide (Ni(OH)_2). (c) The limiting reactant is potassium hydroxide (KOH). (d) The mass of the precipitate formed is 0.9271 grams of Ni(OH)_2. (e) The remaining concentrations in the solution are 0.0667 M Ni^2+, 0.0667 M SO_4^2-, and 0.133 M K^+.

Step-by-step solution

Write the balanced chemical equation for the reaction

The given reactants are potassium hydroxide (KOH) and nickel(II) sulfate (NiSO_{4}). When they react, they swap their anions, forming potassium sulfate (K_2SO_4) and nickel(II) hydroxide (Ni(OH)_2). The balanced chemical equation is: $2\mathrm{KOH}+{\mathrm{NiSO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+\mathrm{Ni}(\mathrm{OH}{)}_2$

Identify the precipitate that forms

In this reaction, nickel(II) hydroxide (Ni(OH)_2) is the precipitate. This compound is insoluble in water, which means that it forms a solid when the two reactants are mixed.

Calculate the moles of each reactant involved

First, we need to find the moles of KOH and NiSO_4 present in the solution. Moles of KOH = Molarity × Volume = 0.200 M × 0.100 L = 0.0200 moles Moles of NiSO_4 = 0.150 M × 0.200 L = 0.0300 moles

Determine the limiting reactant

From the balanced chemical equation, we see that it takes 2 moles of KOH to react with 1 mole of NiSO_4. Let's check if we have enough KOH to react with all of NiSO_4: 0.0200 moles KOH / 2 = 0.0100 moles Since this value (0.0100) is less than 0.0300 moles of NiSO_4, KOH is the limiting reactant.

Calculate the mass of the precipitate

Now that we know the limiting reactant (KOH), we can calculate how many moles of the precipitate (Ni(OH)_2) will form: (0.0200 moles KOH / 2) × 1 mole Ni(OH)_2 = 0.0100 moles Ni(OH)_2 To find the mass of the precipitate, multiply the moles by its molar mass: Mass = 0.0100 moles Ni(OH)_2 × 92.71 g/mol (molar mass of Ni(OH)_2) = 0.9271 g So, 0.9271 grams of Ni(OH)_2 precipitate will form.

Determine the concentration of ions that remain in the solution

Since KOH is the limiting reactant, all of it will be consumed. We need to find out how much NiSO_4 remains unreacted: 0.0300 moles NiSO_4 (initial) - 0.0100 moles (reacted) = 0.0200 moles NiSO_4 (remaining) Now, let's calculate the concentration of the remaining ions in the solution. The total volume of the solution is 100.0 mL + 200.0 mL = 300.0 mL (0.300 L). Concentration of Ni^2+ ions: (0.0200 moles NiSO_4 remaining) / (0.300 L) = 0.0667 M Concentration of SO_4^2- ions: (0.0200 moles NiSO_4 remaining) / (0.300 L) = 0.0667 M Concentration of K^+ ions: (0.0667 M remaining) × 2 = 0.133 M (since 2 moles of K^+ ions are produced per mole of NiSO_4 consumed) The remaining concentrations are 0.0667 M Ni^2+, 0.0667 M SO_4^2-, and 0.133 M K^+.

Key concepts

Chemical Reactions

Let's dive into the fascinating world of chemical reactions. A chemical reaction involves the transformation of one or more substances, known as reactants, into new substances, called products. In the example given, we start with potassium hydroxide (KOH) and nickel(II) sulfate (NiSO₄) as our reactants. These two substances undergo a reaction where they swap their anions. This process results in the formation of potassium sulfate (K₂SO₄) and nickel(II) hydroxide (Ni(OH)₂).

This type of reaction is called a double displacement reaction. The key here is to ensure the equation is balanced. For every reaction, the number of each type of atom on the reactant side should equal the number on the product side. The balanced equation for this reaction is:

• 2 KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

Balancing chemical equations is crucial because it reflects the conservation of mass. We're ensuring that no atoms get lost or created in the process.

Precipitation Reactions

Precipitation reactions occur when two solutions are mixed and an insoluble solid, called a precipitate, forms. In our example, the reaction between KOH and NiSO₄ leads to the formation of a precipitate. The compound Ni(OH)₂ is the precipitate here.

Understanding solubility rules helps us predict when a precipitate will form. Generally, most sulfates are soluble, but there are exceptions for compounds like barium sulfate and nickel hydroxide. Due to its insolubility, Ni(OH)₂ settles out of the solution as a solid.

To visualize this, imagine mixing two clear solutions and seeing a solid form that sinks to the bottom. It's a bit like watching white clouds form in a jar of clear water.

Limiting Reactant

The limiting reactant is the substance that gets completely used up first in a chemical reaction, thus stopping the reaction from continuing because there is nothing left for other reactants to react with. Identifying the limiting reactant helps us determine how much product can be formed.

In this specific reaction, we determine the moles present of each reactant:

• KOH: 0.0200 moles
• NiSO₄: 0.0300 moles

Given that the balanced equation shows a 2:1 ratio (two moles of KOH react with one mole of NiSO₄), we calculate that KOH will run out first. Thus, KOH is our limiting reactant. Once all the KOH has reacted, the process halts as there are no more KOH moles available to react with NiSO₄.

Solution Concentration

Solution concentration conveys how much solute is present in a given volume of solvent, typically expressed in molarity (M), which is moles of solute per liter of solution. After the reaction and removal of the precipitate, we'll need to calculate what's left in the solution.

In this exercise, post-reaction concentrations of ions in solution are found by calculating the leftover reactants and any new products. Given that KOH is the limiting reactant, it is completely consumed. We then calculate the leftover moles of NiSO₄ and the total volume of the solution, converting it as necessary to liters to find concentrations.

• Remaining Ni²⁺ and SO₄²⁻ ions: 0.0667 M
• K⁺ ions: 0.133 M

It's crucial to keep track of volumes and molarity to correctly interpret the reaction's aftermath, especially in precipitation reactions where solids remove ions from the solution.