If 100.0 mL of 0.150 M sodium nitrate is added to 50.0 mL of 0.150 M sodium carbonate, what is the resulting molar concentration of the sodium ion?
mols Na ion from NaNO3 is M x L = ?
mols Na ion from Na2CO3 is 2*M x L = ?
M Na ion is total mols Na ion/total L = ?
To find the resulting molar concentration of the sodium ion, we first need to determine the number of moles of sodium ions present in each solution, then calculate the total number of moles of sodium ions after the solutions are mixed.
To find the number of moles in a solution, we use the formula:
moles = concentration (in M) × volume (in L)
Given that the initial volume of the sodium nitrate solution is 100.0 mL (0.100 L) and the molar concentration is 0.150 M, we can calculate the number of moles of sodium ions in the sodium nitrate solution as follows:
moles of sodium ions in sodium nitrate solution = 0.150 M × 0.100 L
Next, let's calculate the number of moles of sodium ions in the sodium carbonate solution. The initial volume of the sodium carbonate solution is 50.0 mL (0.050 L) and the molar concentration is 0.150 M:
moles of sodium ions in sodium carbonate solution = 0.150 M × 0.050 L
Now, we can sum the number of moles of sodium ions from both solutions to find the total moles of sodium ions:
total moles of sodium ions = moles of sodium ions in sodium nitrate solution + moles of sodium ions in sodium carbonate solution
Finally, to find the resulting molar concentration of the sodium ion, we divide the total moles of sodium ions by the total volume of the solution. The total volume is the sum of the initial volumes of both solutions:
resulting molar concentration of sodium ion = total moles of sodium ions / total volume (in L)
By plugging in the calculated values, we can find the answer to the question.