Which solutions has more sodium (assuming all Na dissociates in all cases) .50 M NaCl or 5%(w/v) CH3COONa)
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I almost answered, but realized this isn't a chemistry question. It's a question about Portland State.
To determine which solution has more sodium, we need to compare the concentrations of sodium in each solution. Let's calculate it step by step:
1. Sodium chloride (NaCl):
A 0.50 M NaCl solution means it contains 0.50 moles of NaCl per liter of solution. Since NaCl disassociates completely into Na+ and Cl- ions, the concentration of sodium ions (Na+) in the solution is also 0.50 M.
2. Sodium acetate (CH3COONa):
A 5%(w/v) CH3COONa solution means it contains 5 grams of CH3COONa per 100 mL of solution.
To determine the concentration of sodium ions, we need to convert the mass of CH3COONa to moles. The molar mass of CH3COONa is:
C = 12.01 g/mol
H = 1.01 g/mol (3 hydrogens)
3 × C + 3 × H + O + Na = 3 × 12.01 + 3 × 1.01 + 16 + 22.99 ≈ 82.03 g/mol
Now, let's convert the mass of CH3COONa to moles:
5 grams CH3COONa * (1 mol / 82.03 g) = 0.061 mol CH3COONa
Since CH3COONa disassociates into CH3COO- and Na+, the concentration of sodium ions (Na+) in the solution is the same as the concentration of CH3COONa, which is 0.061 M.
Comparing the results:
- The concentration of sodium ions in the NaCl solution is 0.50 M.
- The concentration of sodium ions in the CH3COONa solution is 0.061 M.
Therefore, the NaCl solution has a higher concentration of sodium ions compared to the CH3COONa solution.