0.1148 M is the sodium sulfide concentration.
This is the dilution equation: M1 V1 = M2 V2
where M1 is the stock solution's molarity.
V1 is the stock solution's volume.
A diluted solution's M2 molarity.
V2 is the amount of the diluted solution.
Given:
M1 = 0.875 M
V2 = 250 mL and V1 = 50 mL
The ultimate sodium sulfide concentration in V2 must be determined.
Regarding changing values in the equation,
M1V1 = M2V2
M2 = M1V1 ÷ V2
M2 = (0.574 M × 30.0 mL) ÷ 150 mL
M2 = 0.1148 M
The concentration of salt ions in the solution is 2 0.1148 M = 0.2296 M 0.1148 M because one mole containing sodium sulfide produces two moles of sodium ions.
Read more about concentration at
https://brainly.com/question/11471182?referrer=searchResults
#SPJ4
