Respuesta :
Answer:
0.1035 M
Explanation:
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]
Sodium chloride will furnish Sodium ions as:
[tex]NaCl\rightarrow Na^{+}+Cl^-[/tex]
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:
[tex]Moles =0.288 \times {3.58\times 10^{-3}}\ moles[/tex]
Moles of sodium ions by sodium chloride = 0.00103104 moles
Sodium sulfate will furnish Sodium ions as:
[tex]Na_2SO_4\rightarrow 2Na^{+}+SO_4^{2-}[/tex]
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:
[tex]Moles =2\times 0.001 \times {6.51\times 10^{-3}}\ moles[/tex]
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:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity_{Na^+}=\frac{0.00104406}{10.09\times 10^{-3}}[/tex]
The final concentration of sodium anion = 0.1035 M
