Question

Many metal ions form insoluble sulfide compounds when a solution of the metal ion is treated with hydrogen sulfide gas. For example, nickel(II) precipitates nearly quantitatively as NiS when \(\mathrm{H}_{2} \mathrm{S}\) gas is bubbled through a nickel ion solution. How many milliliters of gaseous \(\mathrm{H}_{2} \mathrm{S}\) at STP are needed to precipitate all the nickel ion present in \(10 .\) mL of \(0.050 M \mathrm{NiCl}_{2}\) solution?

Short answer

To precipitate all the nickel ions present in 10 mL of 0.050 M NiCl₂ solution, 11.3 mL of H₂S gas is needed at STP. This is calculated using the stoichiometric ratio between NiCl₂ and H₂S (1:1), the ideal gas law equation, and the conditions of STP.

Step-by-step solution

Write down the balanced chemical equation

The reaction between nickel(II) chloride (NiCl₂) and hydrogen sulfide (H₂S) gas is: \[\ce{NiCl2 + H2S -> NiS(s) + 2HCl}\]

Determine the stoichiometric ratio

From the balanced chemical equation, we can see that 1 mole of NiCl₂ reacts with 1 mole of H₂S. Therefore, the stoichiometric ratio between NiCl₂ and H₂S is 1:1.

Calculate the amount of nickel(II) chloride

We'll calculate the amount (in moles) of nickel(II) chloride (NiCl₂) in the solution using the given concentration and volume: Amount of NiCl₂ (in moles) = concentration × volume = 0.050 M × 10 mL = 0.050 M × 0.01 L = 0.0005 moles

Calculate the required amount of hydrogen sulfide

Using the stoichiometric ratio from Step 2 (1:1), we conclude that the required amount of H₂S is equal to the amount of NiCl₂. Amount of H₂S (in moles) = 0.0005 moles

Convert the required amount of hydrogen sulfide into volume

To calculate the volume (in milliliters) of hydrogen sulfide (H₂S) gas at STP (Standard Temperature and Pressure), we'll use the ideal gas law equation, PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the universal gas constant (0.0821 L atm K⁻¹ mol⁻¹), and T is temperature (in Kelvin). At STP, temperature T = 273.15 K and pressure P = 1 atm. Rearranging the ideal gas law equation, we have: V = (nRT) / P = (0.0005 moles × 0.0821 L atm K⁻¹ mol⁻¹ × 273.15 K) / 1 atm = 0.0113 L Converting the volume to milliliters, we get: Volume of H₂S gas required at STP = 0.0113 L × 1000 mL/L = 11.3 mL Hence, 11.3 mL of H₂S gas is needed at STP to precipitate all the nickel ions present in 10 mL of the 0.050 M NiCl₂ solution.

Key concepts

Balanced Chemical Equations

When carrying out reactions, writing balanced chemical equations is crucial. It demonstrates the conservation of mass and atoms during a chemical reaction. In the exercise, we are precipitating nickel ions using hydrogen sulfide gas. This process starts with the formation of the insoluble nickel sulfide (S(s)).

The balanced equation for the reaction is written as:

• \(\text{NiCl}_2 + \text{H}_2\text{S} \rightarrow \text{NiS(s)} + 2\text{HCl}\)

The equation shows that one molecule of nickel(II) chloride reacts with one molecule of hydrogen sulfide gas to form nickel sulfide and two molecules of hydrochloric acid.

Balancing ensures that the same amount of each atom is present on both sides of the equation. In this reaction, there is a one-to-one ratio between Nickle(II) chloride and hydrogen sulfide, making the stoichiometric calculations straightforward.

Molecular Calculations

In any chemical process, it's important to calculate the moles of reactants accurately. This involves using the molarity and volume of solutions to find the number of moles, which is crucial for stoichiometry.

For our problem, knowing that 10 mL of a 0.050 M \(\text{NiCl}_2\) solution is used, we can find the moles of \(\text{NiCl}_2\) as follows:

• Use the formula: \(\text{Moles} = \text{concentration} \times \text{volume}\)
• Convert mL to L for accuracy: 10 mL = 0.01 L
• Calculate: \(0.050 \ \text{M} \times 0.01 \ \text{L} = 0.0005 \ \text{moles}\)

This tells us there are 0.0005 moles of \(\text{NiCl}_2\) in the solution. Given the 1:1 stoichiometric ratio from the balanced equation, the required moles of\(\text{H}_2\text{S}\) are also 0.0005 moles.

Ideal Gas Law

The Ideal Gas Law is an essential tool for calculating the volume of gases under standard conditions. It relates four properties of a gas: pressure (P), volume (V), amount (), and temperature (T) through the equation \(PV = nRT\).

• \(n\): number of moles
• \(R\): universal gas constant (0.0821 \ \text{L atm K}^{-1} \ \text{mol}^{-1})
• \(T\): temperature in Kelvin (273.15 K at STP)

By rearranging the equation to find volume (V), you can substitute the values:

• \(V = \frac{nRT}{P} = \frac{0.0005 \times 0.0821 \times 273.15}{1}\)

The calculation gives \(0.0113 \ \text{L}\), which converts to 11.3 mL. Thus, you need 11.3 mL of \(\text{H}_2\text{S}\) gas to fully precipitate the nickel ions under standard conditions.