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

Step-by-step 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)} * V_{NaHCO_(3)}`

=
`0.010 * 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)} * V_{Na_(2)CO_(3)}`

=
`0.010 M * 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 * 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}}`

=
`(3 * 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.