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

Explanation:

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

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

Molarity of $NaHCO_3$ solution = 0.010 M

Volume of solution = 10 mL

Putting values in equation 1, we get:

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

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

Thus $10^{-4}mol$ of $NaHCO_3$ contain= $\frac{1}{1}\times 10^{-4}=10^{-4}$ mol of $Na^+$

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

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

Thus $10^{-4}mol$ of $NaHCO_3$ contain= $\frac{2}{1}\times 10^{-4}=2\times 10^{-4}$ mol of $Na^+$

Total $[Na^+]=\frac {\text {total moles}}{\text {total volume}}=\frac{10^{-4}+2\times 10^{-4}}{0.02L}=0.015M$

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