Question

Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate is added to 30.0 mL of 1.0 M sodium bicarbonate.

Answer 1

Answer : The concentration of sodium ion is, 4.5 mole/L

Solution : Given,

Concentration of sodium carbonate = 3 mole/L

Concentration of sodium bicarbonate = 1 mole/L

Volume of sodium carbonate = 70 ml = 0.07 L      (1 L = 1000 ml)

Volume of sodium carbonate = 30 ml = 0.03 L

First we have to calculate the moles of sodium carbonate and sodium bicarbonate.

$\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3mole/L\times 0.07 L=0.21mole$

$\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1mole/L\times 0.03 L=0.03mole$

As we know that there are 2 sodium ion present in sodium carbonate and 1 sodium ion in sodium bicarbonate. Thus, the total moles of sodium ion will be,

$\text{Moles of }Na^+=[2\times (\text{Moles of }Na_2CO_3)+\text{Moles of }NaHCO_3]=2\times 0.21+0.03=0.42+0.03=0.45moles$

Now we have to calculate the concentration of sodium ion.

Total volume = 0.07 + 0.03 = 1 L

$\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Total volume}}=\frac{0.45mole}{0.1L}=4.5mole/L$

Therefore, the concentration of sodium ion is, 4.5 mole/L

Answer 2

The solution would be like this for this specific problem:

 

Molecular formula of the two substances: Na2(CO3) and Na(HCO3)

 

3M * .07L = .21mol 
1M * .03L = .03mol 

.42mol+.03mol = .45mol Na 

30mL + 70mL=100mL or .1L

.45mol/.1L is 4.5M

 

So the sodium ion concentration is 4.5M. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.