Questions: A person suffering from hyponatremia has a sodium ion concentration in the blood of 0.110 M and a total blood volume of 4.50 L. What mass of sodium chloride would need to be added to the blood to bring the sodium ion concentration up to 0.138 M assuming no change in blood volume?

Solution Steps

First, calculate the initial amount of sodium ions in the blood. The initial concentration is \(0.110 \, \text{M}\) and the blood volume is \(4.50 \, \text{L}\).

\[ \text{Initial moles of Na}^+ = 0.110 \, \text{mol/L} \times 4.50 \, \text{L} = 0.495 \, \text{mol} \]

Next, calculate the desired amount of sodium ions for a concentration of \(0.138 \, \text{M}\).

\[ \text{Desired moles of Na}^+ = 0.138 \, \text{mol/L} \times 4.50 \, \text{L} = 0.621 \, \text{mol} \]

Subtract the initial moles of sodium ions from the desired moles to find the additional moles needed.

\[ \text{Additional moles of Na}^+ = 0.621 \, \text{mol} - 0.495 \, \text{mol} = 0.126 \, \text{mol} \]

Sodium chloride (NaCl) dissociates into sodium ions (Na\(^+\)) and chloride ions (Cl\(^-\)) in solution. Therefore, 1 mole of NaCl provides 1 mole of Na\(^+\).

The molar mass of NaCl is approximately \(58.44 \, \text{g/mol}\).

\[ \text{Mass of NaCl} = 0.126 \, \text{mol} \times 58.44 \, \text{g/mol} = 7.358 \, \text{g} \]

Final Answer