Question

A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution

Short answer

The concentration of the NaOH solution is 0.0130 M.

Step-by-step solution

Set up the given information and list the formulas needed

We're given that 50.0 mL of an NaOH solution is added to 100.0 mL of a 0.400 M HCl solution. Also, the solution forms 2.06 g of precipitate when treated with chromium(III) nitrate. First, let's list the relevant chemical reactions: 1. Neutralization reaction between NaOH and HCl: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) 2. Reaction between remaining OH- ions and Cr(NO3)3: 3OH^(-)(aq) + Cr(NO3)3(aq) -> Cr(OH)3(s) + 3NO3^(-)(aq) The molar mass of Cr(OH)3 is: 51.996 (Cr) + 3(15.999) (O) + 3(1.008) (H) = 151.994 g/mol Mole calculation formula for given mass: n = mass / molar mass The concentration formula: C = n / V

Calculate the moles of precipitate formed and OH- ions left

To determine the number of moles of Cr(OH)3 precipitate formed, we need to divide the mass of Cr(OH)3 by its molar mass: n(Cr(OH)3) = (mass of Cr(OH)3) / (molar mass of Cr(OH)3) n(Cr(OH)3) = 2.06 g / 151.994 g/mol = 0.01355 mol From the stoichiometry of the reaction between OH- ions and Cr(NO3)3, 1 unit of Cr(OH)3 precipitate is formed by reacting 3 units of OH- ions. So, we'll multiply the moles of precipitate by 3 to find moles of OH- ions. n(OH-) = 3 * n(Cr(OH)3) n(OH-) = 3 * 0.01355 mol = 0.04065 mol

Calculate the moles of HCl and NaOH initially and find the concentration of NaOH solution

Moles of HCl = (volume of HCl solution) * (concentration of HCl solution) n(HCl) = 100.0 mL * 0.400 mol/L = 0.0400 mol (considering volume in L, 100.0 mL = 0.100 L) Initially, the moles of NaOH needed to neutralize the given moles of HCl will be equal to the moles of HCl (as in the neutralization reaction, one mole of NaOH reacts with one mole of HCl). Subtracting these moles from the total moles of OH- ions will give us the moles of NaOH in 50.0 mL. n(NaOH) = n(OH-) - n(HCl) n(NaOH) = 0.04065 mol - 0.0400 mol = 0.00065 mol Now, let's find the concentration of the NaOH solution using the moles of NaOH and volume of NaOH solution: C(NaOH) = n(NaOH) / V(NaOH) C(NaOH) = 0.00065 mol / 0.050 L = 0.0130 M So, the concentration of the NaOH solution is 0.0130 M.

Key concepts

Neutralization Reaction

In chemistry, a neutralization reaction occurs when an acid and a base react to form water and a salt. This is a fundamental type of chemical reaction. The essential part of the reaction is the combination of the H^+ ions from the acid and the OH^- ions from the base to create water. For example, when hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), it forms sodium chloride (NaCl) and water (H2O). Here is how the reaction looks:
\[\text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}\]This equation shows that one mole of NaOH reacts with one mole of HCl to produce one mole of NaCl and one mole of water.

• Acid: Donates H^+ ions
• Base: Donates OH^- ions
• Product: Salt and Water

Understanding this reaction is central to solving problems involving neutralization, such as determining unused reactants or product formation.

Stoichiometry

Stoichiometry is the area of chemistry that involves using balanced chemical equations to calculate relationships between the amounts of reactants and products. It is founded on the conservation of mass and the concept that all reactants in a reaction have a specific ratio, based on the balanced equation. In the context of the given problem, stoichiometry helps us understand the proportions of HCl reacting with NaOH, and subsequently, how the remaining OH^- ions react with Cr(NO3)3 to form Cr(OH)3.
For instance, the reaction:
\[3\text{OH}^- + \text{Cr(NO}_3\text{)}_3 \rightarrow \text{Cr(OH)}_3(\text{s}) + 3\text{NO}_3^-\]indicates that three moles of OH^- are needed to form one mole of Cr(OH)3. This relationship allows us to determine how much OH^- is leftover in the solution after the neutralization reaction with HCl, crucial in calculating the products of subsequent reactions.

• Uses balanced equations
• Predicts reactant and product quantities
• Utilizes molecular ratios

Mole Calculation

Calculating moles is a key process in chemistry, used to quantify the amount of a substance. The mole is a standard unit in chemistry that represents a quantity of chemical entities such as atoms, molecules, or ions. To find moles, you can use the formula:
\[n = \frac{\text{mass}}{\text{molar mass}}\]For the given problem, we use this formula to calculate the moles of Cr(OH)3 formed by reacting with chromium(III) nitrate:
\[n(\text{Cr(OH)}_3) = \frac{2.06 \text{ g}}{151.994 \text{ g/mol}} = 0.01355 \text{ mol}\]Using mole calculations, we then determine the number of excess OH^- ions by considering the stoichiometry of the reactions involved.

• Converts mass to moles
• Essential for reaction calculations
• Link between mass and count of entities

Precipitate Formation

Precipitate formation refers to the creation of a solid from a solution during a chemical reaction. In the problem, a precipitate is formed when OH^- ions react with chromium(III) nitrate to produce chromium(III) hydroxide (Cr(OH)3). Precipitates are typically insoluble solids formed from solute particles in the reactants.
The specific reaction at hand is:
\[3\text{OH}^- + \text{Cr(NO}_3\text{)}_3 \rightarrow \text{Cr(OH)}_3(\text{s}) + 3\text{NO}_3^-\]This equation demonstrates the precipitation process, where the compound Cr(OH)3 settles out of the aqueous solution. The fact that it forms a solid makes it a precipitate.

• Indicates formation of a solid
• Involves ionic compounds
• Important in qualitative analysis

This process is essential in many analytical procedures to isolate and identify different ions in a solution.