The sodium ion concentration in the solution of 17.5 mL of 0.1050 M Na₂CO₃ and 46.0 mL of 0.1250 M NaCl is 0.148 M.
To find the Na⁺ concentration in the final solution, we need to find the number of moles of the ion in the Na₂CO₃ and NaCl solutions and the volume of the final solution.
[tex] C_{Na^{+}} = \frac{n_{t}}{V_{t}} [/tex] (1)
Where:
[tex] n_{t} [/tex]: is the total number of moles of Na⁺
[tex] V_{t} [/tex]: is the volume of the final solution
The number of moles of Na₂CO₃ can be calculated as follows:
[tex] n_{Na_{2}CO_{3}} = C_{Na_{2}CO_{3}}*V_{Na_{2}CO_{3}} [/tex]
Where:
[tex] C_{Na_{2}CO_{3}} [/tex]: is the concentration = 0.1050 mol/L
[tex] V_{Na_{2}CO_{3}} [/tex]: is the volume = 17.5 mL = 0.0175 L
[tex] n_{Na_{2}CO_{3}} = 0.1050 mol/L*0.0175 L = 1.84 \cdot 10^{-3} \:moles [/tex]
We can find the number of moles of Na⁺ knowing that in 1 mol of Na₂CO₃ we have 2 moles of Na⁺.
[tex] n_{Na^{+}} = \frac{2 \:mol\: Na^{+}}{1\:mol \:Na_{2}CO_{3}}*mol \:Na_{2}CO_{3} = \frac{2 \:mol\: Na^{+}}{1\:mol \:Na_{2}CO_{3}}*1.84 \cdot 10^{-3} \:moles = 3.68 \cdot 10^{-3} \:moles [/tex]
The number of moles of Na⁺ in the NaCl solution is:
[tex] n_{NaCl} = C_{NaCl}*V_{NaCl} = 0.1250 mol/L*0.046 L = 5.75 \cdot 10^{-3} \:moles [/tex]
In 1 mol of NaCl we have 1 mol of Na⁺, so the number of moles of Na⁺ in the NaCl solution is 5.75x10⁻³ moles.
The total number of moles of Na⁺ in the final solution can be calculated as follows
[tex] n_{t} = n_{{Na^{+}}_{Na_{2}CO_{3}}} + n_{{Na^{+}}_{NaCl}} = 3.68 \cdot 10^{-3} \:moles + 5.75 \cdot 10^{-3} \:moles = 9.43 \cdot 10^{-3} moles [/tex]
Hence, the number of moles of Na⁺ in the final solution is 9.43x10⁻³ moles.
The volume of the final solution is given by:
[tex] V_{t} = V_{Na_{2}CO_{3}} + V_{NaCl} = 0.0175 L + 0.046 L = 0.0635 L [/tex]
The concentration of Na⁺ in the final solution is (eq 1):
[tex] C_{Na^{+}} = \frac{n_{t}}{V_{t}} = \frac{9.43 \cdot 10^{-3} moles}{0.0635 L} = 0.148 mol/L [/tex]
Therefore, the concentration of sodium ions in the final solution is 0.148 M.
Find more about molar concentration here:
I hope it helps you!
