Question

In the laboratory a student combines 23.5 mL of a 0.173 M barium sulfide solution with 27.7 mL of a 0.402 M sodium sulfide solution. What is the final concentration of sulfide anion ?

Answer 1

Answer: The final concentration of sulfide anion in the solution is 0.297 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$    .....(1)

• For barium sulfide:

Molarity of barium sulfide solution = 0.173 M

Volume of solution = 23.5 mL

Putting values in equation 1, we get:

$0.173M=\frac{\text{Moles of barium sulfide}\times 1000}{23.5}\\\\\text{Moles of barium sulfide}=\frac{0.173\times 23.5}{1000}=0.0041mol$

1 mole of barium sulfide (BaS) produces 1 mole of barium ions $(Ba^{2+})$ and 1 mole of sulfide ions $(S^{2-})$

Moles of sulfide ions = (1 × 0.0041) = 0.0041 moles

• For sodium sulfide:

Molarity of sodium sulfide solution = 0.402 M

Volume of solution = 27.7 mL

Putting values in equation 1, we get:

$0.402M=\frac{\text{Moles of sodium sulfide}\times 1000}{27.7}\\\\\text{Moles of sodium sulfide}=\frac{0.402\times 27.7}{1000}=0.0111mol$

1 mole of sodium sulfide $(Na_2S)$ produces 2 moles of sodium ions $(Na^{+})$ and 1 mole of sulfide ions $(S^{2-})$

Moles of sulfide ions = (1 × 0.0111) = 0.0111 moles

Now, calculating the concentration of sulfide anion in the solution by using equation 1, we get:

Total moles of sulfide ions = [0.0041 + 0.0111] = 0.0152 moles

Total volume of the solution = [23.5 + 27.7]  = 51.2 mL

Putting values in equation 1, we get:

$\text{Molarity of }S^{2-}\text{ ions}=\frac{0.0152\times 1000}{51.2}\\\\\text{Molarity of }S^{2-}\text{ ions}=0.297M$

Hence, the final concentration of sulfide anion in the solution is 0.297 M