Respuesta :
Explanation:
Total volume of the solution will be (10 ml + 10 ml) = 20 ml. As there are 1000 ml in 1 L. So, 20 ml will be equal to 0.02 L.
As molarity equals number of moles divided by volume in liter. Hence, calculate number of moles of [tex]NaHCO_{3}[/tex] as follows.
No. of moles = [tex]M_{NaHCO_{3}} \times V_{NaHCO_{3}}[/tex]
= [tex]0.010 \times 0.01 L[/tex]
= [tex]10^{-4}[/tex] mol
Hence, moles of [tex]Na^{+}[/tex] will also be equal to [tex]10^{-4}[/tex] mol.
Now, calculate the no. of moles of [tex]Na_{2}CO_{3}[/tex] as follows.
No. of moles = [tex]M_{Na_{2}CO_{3}} \times V_{Na_{2}CO_{3}}[/tex]
= [tex]0.010 M \times 0.01 L[/tex]
= [tex]10^{-4}[/tex] mol
As there are 2 [tex]Na^{+}[/tex] ions in 1 mole of [tex]Na_{2}CO_{3}[/tex]. Hence, in [tex]10^{-4}[/tex] mol number of [tex]Na^{+}[/tex] ions will be [tex]2 \times 10^{-4}[/tex] mol.
When both the solutions are mixed together then molarity of the solution will be calculated as follows.
Molarity of solution = [tex]\frac{total Na^{+} ions}{\text{total volume}}[/tex]
= [tex]\frac{3 \times 10^{-4}}{0.02L}[/tex]
= 0.015 M
Thus, we can conclude that the molar concentration of Na (aq) in the given solution is 0.015 M.
