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?

A. 0.010 mole/L
B. 0.015 mole/L
C. 0.020 mole/L
D. 0.030 mole/L

Answer 1

Answer: B. 0.015 mole/L

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in L)}}` .....(1)

Molarity of
`NaHCO_3` solution = 0.010 M

Volume of solution = 10 mL

Putting values in equation 1, we get:

a)
`0.010M=\frac{\text{Moles of}NaHCO_3* 1000}{10ml}\\\\\text{Moles of }NaHCO_3=(0.010mol/L* 10)/(1000)=10^(-4)mol`

1 mole of
`NaHCO_3` contains = 1 mol of
`Na^+`

Thus
`10^(-4)mol` of
`NaHCO_3` contain=
`(1)/(1)* 10^(-4)=10^(-4)` mol of
`Na^+`

b)
`0.010M=\frac{\text{Moles of}Na_2CO_3* 1000}{10ml}\\\\\text{Moles of }Na_2CO_3=(0.010mol/L* 10)/(1000)=10^(-4)mol`

1 mole of
`Na_2CO_3` contains = 2 mol of
`Na^+`

Thus
`10^(-4)mol` of
`NaHCO_3` contain=
`(2)/(1)* 10^(-4)=2* 10^(-4)` mol of
`Na^+`

Total
`[Na^+]=\frac {\text {total moles}}{\text {total volume}}=(10^(-4)+2* 10^(-4))/(0.02L)=0.015M`

The molar concentration of
`Na^+` in a solution is 0.015 M