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.
[tex]\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[/tex]
[tex]\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1mole/L\times 0.03 L=0.03mole[/tex]
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,
[tex]\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[/tex]
Now we have to calculate the concentration of sodium ion.
Total volume = 0.07 + 0.03 = 1 L
[tex]\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Total volume}}=\frac{0.45mole}{0.1L}=4.5mole/L[/tex]
Therefore, the concentration of sodium ion is, 4.5 mole/L
The concentration of sodium ions in the mixture is 4.5 M.
Number of moles of sodium bicarbonate = 30/1000 × 1.0 M = 0.03 moles
We know that there is only one mole of sodium ions in sodium bicarbonate (NaHCO3).
Number of moles in sodium bicarbonate = 0.03 moles
Number of moles of sodium carbonate = 70/1000 × 3.0 M = 0.21 moles
We know that there are two moles of sodium in sodium carbonate (Na2CO3)
Number of moles of sodium in sodium carbonate = 2 × 0.21 moles = 0.42 moles of sodium carbonate
Total volume of solution = 70.0 mL + 30.0 mL = 100 mL or 0.1 L
Total number of moles of sodium in the mixture = 0.03 moles + 0.42 = 0.45 moles
Total concentration of sodium ions in the mixture = 0.45 moles/0.1 L
= 4.5 M
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