Calculate the sodium ion concentration when \(70.0 \mathrm{mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

We are given the following information: - Sodium carbonate solution: \(70.0 \mathrm{mL}\) volume and \(3.0 \mathrm{M}\) concentration (note that one mole of sodium carbonate \(\mathrm{Na_2CO_3}\) contains two moles of sodium ions \(\mathrm{Na^+}\)) - Sodium bicarbonate solution: \(30.0 \mathrm{mL}\) volume and \(1.0 \mathrm{M}\) concentration (one mole of sodium bicarbonate \(\mathrm{NaHCO_3}\) contains one mole of sodium ions \(\mathrm{Na^+}\)) To find the moles of sodium ions in each solution we use the formula: moles = volume × concentration. Note we need to multiply with the number of ions in each molecule for Sodium carbonate. Moles of sodium ions in sodium carbonate solution: \(2 \times (70.0 \mathrm{mL} \times 3.0 \mathrm{M}) = 2 \times (210.0 \mathrm{mmol}) = 420.0 \mathrm{mmol}\) Moles of sodium ions in sodium bicarbonate solution: \(30.0 \mathrm{mL} \times 1.0 \mathrm{M} = 30.0 \mathrm{mmol}\)

Next, we need to find the total volume of the final solution. To do this, simply add the volumes of each solution together: Total volume = volume of sodium carbonate + volume of sodium bicarbonate Total volume = \(70.0 \mathrm{mL} + 30.0 \mathrm{mL} = 100.0 \mathrm{mL}\)

Now, we need to find the total moles of sodium ions in the final solution. To do this, add the moles of sodium ions from each solution: Total moles = moles of sodium ions in sodium carbonate + moles of sodium ions in sodium bicarbonate Total moles = \(420.0 \mathrm{mmol} + 30.0 \mathrm{mmol} = 450.0 \mathrm{mmol}\)

Finally, we can find the sodium ion concentration in the final solution by dividing the total moles of sodium ions by the total volume of the solution: Sodium ion concentration = \(\frac{\text{Total moles}}{\text{Total volume}}\) Sodium ion concentration = \(\frac{450.0 \mathrm{mmol}}{100.0 \mathrm{mL}} = 4.50 \mathrm{M}\) Therefore, the sodium ion concentration in the final solution is \(4.50 \mathrm{M}\).