Question

Calculate the concentration (M) of sodium ions in a solution made by diluting 10.0 mL of a 0.774 M solution of sodium sulfide to a total volume of 100 mL.

Answer 1

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.

Dilution

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:

$m_1v_1 = m_2v_2$

Where $m_1$ is the molarity before dilution, $v_1$ is the volume before dilution, $m_2$ is the molarity after dilution, and $v_2$ is the volume after dilution.

In this case, $m_1$ = 0.774 M, $v_1$ = 10.0 mL, $v_2$ = 100 mL.

$m_2$ = $m_1v_1/v_2$

     = 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 = $Na_2S$ ---> $2Na^+ + S^{2-$

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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