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 ?

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:

   .....(1)

Molarity of barium sulfide solution = 0.173 M

Volume of solution = 23.5 mL

Putting values in equation 1, we get:

1 mole of barium sulfide (BaS) produces 1 mole of barium ions and 1 mole of sulfide ions

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

Molarity of sodium sulfide solution = 0.402 M

Volume of solution = 27.7 mL

Putting values in equation 1, we get:

1 mole of sodium sulfide produces 2 moles of sodium ions and 1 mole of sulfide ions

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:

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

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2