Calculate the sodium ion concentration when 50.0 mL of 2.0 M sodium carbonate is added to 100.0 mL of 3.0 M sodium bicarbonate.
mols Na2CO3 = M x L = ?
mols Na^+ = twice that since there are two Na ions in 1 molecule of Na2CO3.
mols NaHCO3 = M x L = ?
Total Na = mols Na from Na2CO3 + mols Na from NaHCO3.
M Na = mols Na/total volume in L
To calculate the sodium ion concentration in the final solution, we need to first determine the number of moles of sodium ions coming from each compound and then use that information to calculate the total concentration.
Let's start by calculating the number of moles of sodium ions coming from sodium carbonate:
Molarity (M) = moles (mol) / volume (L)
Given:
Molarity of sodium carbonate = 2.0 M
Volume of sodium carbonate = 50.0 mL = 0.050 L
Using the formula: moles = Molarity * volume
moles of sodium carbonate = 2.0 M * 0.050 L = 0.100 mol
Now let's calculate the number of moles of sodium ions coming from sodium bicarbonate:
Molarity of sodium bicarbonate = 3.0 M
Volume of sodium bicarbonate = 100.0 mL = 0.100 L
moles of sodium bicarbonate = 3.0 M * 0.100 L = 0.300 mol
Since sodium carbonate (Na2CO3) and sodium bicarbonate (NaHCO3) both produce two moles of sodium ions (Na+) per mole, we need to double the number of moles we obtained for each compound.
Total moles of sodium ions = (2 * moles of sodium carbonate) + (2 * moles of sodium bicarbonate)
Total moles of sodium ions = (2 * 0.100 mol) + (2 * 0.300 mol)
Total moles of sodium ions = 0.200 mol + 0.600 mol
Total moles of sodium ions = 0.800 mol
To find the concentration, we divide the total moles of sodium ions by the total volume of the final solution, which is the sum of the volumes of sodium carbonate and sodium bicarbonate.
Total volume = volume of sodium carbonate + volume of sodium bicarbonate
Total volume = 0.050 L + 0.100 L
Total volume = 0.150 L
Finally, we can calculate the sodium ion concentration:
Sodium ion concentration = Total moles of sodium ions / Total volume
Sodium ion concentration = 0.800 mol / 0.150 L
Sodium ion concentration = 5.33 M
Therefore, the sodium ion concentration in the final solution is 5.33 M.