Which solutions has more sodium (assuming all Na dissociates in all cases) .50 M NaCl or 5%(w/v) CH3COONa)
Your question begs another question. What exactly do you mean by "has more sodium"?
Doesn't it depend upon how much of the NaCl and CH3COONa solutions we compare?
this is why I posted this question it is a question that my professor gave and I am stumped
To determine which solution has more sodium, we need to compare the molar concentrations of the sodium ions in each solution.
In the case of sodium chloride (NaCl), when it dissociates in water, it forms one sodium ion (Na+) and one chloride ion (Cl-). Since the solution is 0.50 M NaCl, the concentration of sodium ions will also be 0.50 M.
On the other hand, for sodium acetate (CH3COONa), when it dissociates in water, it releases one sodium ion (Na+), one acetate ion (CH3COO-), and one hydrogen ion (H+). However, only the sodium ion contributes to the sodium concentration. The 5%(w/v) concentration means that 5 grams of sodium acetate are dissolved in 100 mL of water. To calculate the molar concentration, we need to convert grams to moles and volume to liters.
The molar mass of sodium acetate (NaC2H3O2) is approximately 82.03 g/mol. So, 5 grams of sodium acetate is equal to 5/82.03 = 0.061 mol.
Since we have 0.061 mol of sodium acetate dissolved in a volume of 100 mL (which is 0.1 L), we can calculate the molar concentration:
Concentration (molarity) = moles/volume
Concentration = 0.061 mol / 0.1 L = 0.61 M
Therefore, the molar concentration of sodium ions in the 5%(w/v) CH3COONa solution is 0.61 M.
Comparing the two solutions, we find that the concentration of sodium ions is higher in the 0.50 M NaCl solution compared to the 0.61 M sodium ions in the 5%(w/v) CH3COONa solution.
Hence, the 0.50 M NaCl solution has a higher concentration of sodium ions.