By taking into account the different numbers of sodium ions released by different salts upon dissociation, we can calculate that the final molarity of sodium salts is 0.1055 M.
To calculate the total molarity of the sodium ions in the new solution, we must calculate the number of moles of sodium ions in the first, second, and third solutions, add them up and divide them by the total volume of the new solution.
First solution - 1 mol of sodium chloride releases 1 mole of sodium ions upon dissociation, so the amount of sodium ions will be equal to the amount of sodium chloride:
c = n/V ⇒ n = c*V = 1.25 M * 0.0250 L = 0.03125 mol
Second solution - 1 mol of sodium sulfate releases 2 moles of sodium ions upon dissociation, so the amount of sodium ions will be double the amount of sodium sulfate:
n = 2c*V = 2 * 0.550 M * 0.145 L = 0.1595 mol
Third solution - 1 mol of sodium phosphate releases 3 moles of sodium ions upon dissociation, so the amount of sodium ions will be triple the amount of sodium phosphate:
n = 3c*V = 3 * 0.225 M * 0.0300 L = 0.02025 mol
The final molarity of sodium ions:
c = n/V = (0.03125 mol + 0.1595 mol + 0.02025 mol) / (0.0250 L + 0.145 L + 0.0300 L) = 1.055 M
You can learn more about dissociation here:
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