Solid sodium sulfide is slowly added to 55.0 mL of a calcium iodide solution until the concentration of sulfide ion is 0.0396M. The maximum amount of calcium ion remaining in solution is ______ M.
It's not given in the question :(
Always remember to take all the given quantities in SI unit.
Maximum amount of calcium ion is equal to sulphide ion concentration as in solution we will have one [ca2+] , one [S2-], two [Na+] and two [I-] ion will be present
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