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The concentration of sodium ions in a solution made by diluting 10.0 mL of a 0.774 M solution of sodium sulfide to a volume of 100 mL would be 0.1548 M.
<h3>Dilution</h3>According to the dilution principle, the number of moles of solutes in a solution before and after dilution must be the same. This is expressed as the following equation:
Where is the molarity before dilution,
is the volume before dilution,
is the molarity after dilution, and
is the volume after dilution.
In this case, = 0.774 M,
= 10.0 mL,
= 100 mL.
=
= 0.774 x 10/100
= 0.0774 M
Thus, the new concentration of the sodium sulfide solution would be 0.0774 M.
Mole of the final solution: 0.0774 x 0.1 = 0.00774 mol
Sodium sulfide formula = --->
Equivalent mole of sodium ion = 0.00774 x 2
= 0.01548 mol
The concentration of sodium ions = mol/volume
= 0.01548/0.1
= 0.1548 M
In other words, the concentration of the sodium ions in the diluted solution would be 0.1548 M.
More on dilution can be found here: brainly.com/question/21323871
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