Question

What is the molar concentration of Na (aq) in a solution that is prepared by mixing 10 mL of a 0.010 M NaHCO3(aq) solution with 10 mL of a 0.010 M Na2CO3(aq) solution

Answer 1

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 $NaHCO_{3}$ as follows.

        No. of moles = $M_{NaHCO_{3}} \times V_{NaHCO_{3}}$

                              = $0.010 \times 0.01 L$

                              = $10^{-4}$ mol

Hence, moles of $Na^{+}$ will also be equal to $10^{-4}$ mol.

Now, calculate the no. of moles of $Na_{2}CO_{3}$ as follows.

         No. of moles = $M_{Na_{2}CO_{3}} \times V_{Na_{2}CO_{3}}$

                               = $0.010 M \times 0.01 L$

                               = $10^{-4}$ mol

As there are 2 $Na^{+}$ ions in 1 mole of $Na_{2}CO_{3}$. Hence, in $10^{-4}$ mol number of $Na^{+}$ ions will be $2 \times 10^{-4}$ mol.

When both the solutions are mixed together then molarity of the solution will be calculated as follows.

                 Molarity of solution = $\frac{total Na^{+} ions}{\text{total volume}}$

                                = $\frac{3 \times 10^{-4}}{0.02L}$

                                = 0.015 M  

Thus, we can conclude that the molar concentration of Na (aq) in the given solution is 0.015 M.