Calculate the molarity of sodium ion in a solution made
bymixing 3.58 ml of 0.288 m sodium chloride with 500 ml of 6.51
times1/1000 m sodium sulfate ( assume volumes are additive ).

0.1035 M

Explanation:

Considering:

Sodium chloride will furnish Sodium ions as:

Given :

For Sodium chloride :

Molarity = 0.288 M

Volume = 3.58 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 3.58×10⁻³ L

Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:

Moles of sodium ions by sodium chloride = 0.00103104 moles

Sodium sulfate will furnish Sodium ions as:

Given :

For Sodium sulfate :

Molarity = 0.001 M

Volume = 6.51 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.51 ×10⁻³ L

Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:

Moles of sodium ions by Sodium sulfate = 0.00001302 moles

Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles

Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L

Concentration of sodium ions is:

The final concentration of sodium anion = 0.1035 M

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explanation: