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Alexxx
9 days ago
15

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.

Chemistry
2 answers:
Anarel [852]9 days ago
7 0

Answer: Sodium ion concentration is 4.5 mole/L

Solution: Here are the details,

Sodium carbonate concentration is 3 mole/L

Sodium bicarbonate concentration is 1 mole/L

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

Volume of sodium bicarbonate is 30 ml = 0.03 L

We first need to determine the moles of both 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

Since sodium carbonate contains 2 sodium ions and sodium bicarbonate has 1 sodium ion, the total moles of sodium ions add up to,

\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

Next, we calculate the sodium ion concentration.

Total volume = 0.07 + 0.03 = 0.1 L

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

Consequently, the sodium ion concentration is 4.5 mole/L

castortr0y [927]9 days ago
5 0

The resolution for this specific case will be as follows:

 

The molecular formulas of the two compounds are: Na2(CO3) and Na(HCO3)

 

<span>3M *.07L =.21mol </span><span>
<span>1M *.03L =.03mol </span></span>

<span>.21mol +.03mol =.45mol Na </span>

Mixing together 30mL and 70mL results in a total volume of 100mL or.1L

Calculating the concentration gives.45mol/.1L = 4.5M

 

Thus, the concentration of sodium ions amounts to 4.5M. I trust this answer meets your needs, and don't hesitate to reach out with more questions if needed.

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