Question

When 70. milliliter of 3.0-molar Na2CO3 is added to 30. milliliters of 1.0-molar NaHCO3 the result­ing concentration of Na+ is 2.0 M

Answer 1

Answer : The resulting concentration of $Na^+$ ion is, 4.5 M

Explanation : Given,

Concentration of $Na_2CO_3$ = $M_1$ = 3.0 M = 3.0 mol/L

Volume of $Na_2CO_3$ = $V_1$ = 70 mL = 0.07 L

Concentration of $NaHCO_3$ = $M_2$ = 1.0 M = 1.0 mol/L

Volume of $NaHCO_3$ = $V_2$ = 30 mL = 0.03 L

First we have to calculate the moles of $Na_2CO_3$ and $NaHCO_3$

$\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3.0mol/L\times 0.07L=0.21mol$

and,

$\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1.0mol/L\times 0.03L=0.03mol$

Now we have to calculate the moles of $Na^+$ ions.

As, 1 mole of $Na_2CO_3$ will give 2 moles of $Na^+$ ions

So, 0.21 moles of $Na_2CO_3$ will give $2\times 0.21=0.42$ moles of $Na^+$ ions

and,

As, 1 mole of $NaHCO_3$ will give 1 mole of $Na^+$ ions

So, 0.03 moles of $NaHCO_3$ will give 0.03 moles of $Na^+$ ions

So,

Total number of moles of $Na^+$ ions = 0.42 + 0.03 =0.45 mole

Total volume of both solution = 70 mL + 30 mL = 100 mL = 0.1 L

Now we have to calculate the concentration of $Na^+$ ions.

$\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Volume of solution}}=\frac{0.45mol}{0.1L}=4.5mol/L=4.5M$

Therefore, the resulting concentration of $Na^+$ ion is, 4.5 M