Question

Using modern analytical techniques, it is possible to detect sodium ions in concentrations as low as \(50 \mathrm{pg} / \mathrm{mL}\). What is this detection limit expressed in (a) molarity of \(\mathrm{Na}^{+}\), (b) \(\mathrm{Na}^{+}\) ions per cubic centimeter?

Short answer

The detection limit of sodium ions expressed in (a) molarity of Na+ ions is approximately \(\frac{50}{22.99 \times 10^9}\, \text{mol/L}\) and in (b) Na+ ions per cubic centimeter is approximately \(\frac{50 \times 6.022 \times 10^{23}}{22.99 \times 10^6}\, \text{ions}\).

Step-by-step solution

Convert picograms to grams

Given that the detection limit is 50 pg/mL, we should first convert it to grams per milliliter (g/mL) by dividing by 10^12 (since 1 g = 10^12 pg): \(50\,\text{pg/mL}\) × \(\frac{1\,\text{g}}{10^{12}\,\text{pg}}\) = \(\frac{50}{10^{12}}\,\text{g/mL}\)

Convert milliliters to liters

Next, to convert the concentration from g/mL to g/L, we need to multiply by 1000 (since 1 L = 1000 mL): \(\frac{50}{10^{12}}\,\text{g/mL}\) × \(1000\,\text{mL/L}\) = \(\frac{50}{10^{9}}\,\text{g/L}\)

Calculate the molarity of Na+ ions

To find the molarity, we can divide the mass of sodium ions by the molecular weight of sodium (Na) which is 22.99 g/mol and then divide by the volume in liters: \(\text{Molarity}\) = \(\frac{\text{moles of solute}}{\text{liters of solution}}\) = \(\frac{(\frac{50}{10^9}\, \text{g/L})}{22.99\, \text{g/mol}}\) = \(\frac{\frac{50}{10^9}}{22.99}\, \text{mol/L}\) = \(\frac{50}{22.99 \times 10^9}\, \text{mol/L}\) #a) Answer# The detection limit expressed in molarity of Na+ ions is approximately \(\frac{50}{22.99 \times 10^9}\, \text{mol/L}\). #b) Finding Na+ ions per cubic centimeter#

Calculate the number of moles in one cubic centimeter

We know the concentration is \(\frac{50}{22.99 \times 10^9}\, \text{mol/L}\). To convert it to moles per cubic centimeter, we can multiply by 0.001 L, as 1 cubic centimeter (cc) equals 1 milliliter (mL) and there are 1000 milliliters in a liter: \(\frac{50}{22.99 \times 10^9}\, \text{mol/L}\) × \(0.001\, \text{L}\) = \(\frac{50}{22.99 \times 10^6}\, \text{mol}\)

Convert the moles to number of ions

To find the number of ions, we can multiply the moles by Avogadro's number (\(6.022 \times 10^{23}\,\text{ions/mol}\)): \(\frac{50}{22.99 \times 10^6}\, \text{mol}\) × \(6.022 \times 10^{23}\, \text{ions/mol}\) = \(\frac{50 \times 6.022 \times 10^{23}}{22.99 \times 10^6}\, \text{ions}\) #b) Answer# The detection limit expressed in Na+ ions per cubic centimeter is approximately \(\frac{50 \times 6.022 \times 10^{23}}{22.99 \times 10^6}\, \text{ions}\).