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Chapter 4: Problem 64

A person suffering from hyponatremia has a sodium ion concentration in the blood of \(0.118 \mathrm{M}\) and a total blood volume of \(4.6 \mathrm{~L}\). What mass of sodium chloride would need to be added to the blood to bring the sodium ion concentration up to \(0.138 \mathrm{M}\), assuming no change in blood volume?

Short Answer

Expert verified
Approximately \(5.375 \, \mathrm{g}\) of sodium chloride should be added to the blood to increase the sodium ion concentration to \(0.138 \, \mathrm{M}\) without any change in blood volume.

Step by step solution

01

Calculate the initial amount of sodium ions in the blood

To calculate the initial amount of sodium ions in the blood, we can use the formula: Initial amount of sodium ions = Initial concentration × Blood volume Let's plug in the given values: Initial concentration = 0.118 M Blood volume = 4.6 L Initial amount of sodium ions = \(0.118 \mathrm{M} \times 4.6 \mathrm{L}\) = \(0.5428 \,\text{moles of sodium ions}\)
02

Calculate the final amount of sodium ions required for the desired concentration

To calculate the final amount of sodium ions for a concentration of 0.138 M, we can use the same formula as in step 1: Final amount of sodium ions = Final concentration × Blood volume We have: Final concentration = 0.138 M Blood volume = 4.6 L Final amount of sodium ions = \(0.138 \,\mathrm{M} \times 4.6 \,\mathrm{L}\) = \(0.6348 \,\text{moles of sodium ions}\)
03

Calculate the net amount of sodium ions needed

Now, let's calculate the difference between the initial and final amount of sodium ions: Net amount of sodium ions needed = Final amount of sodium ions - Initial amount of sodium ions Net amount of sodium ions needed = \(0.6348 \,\text{moles} - 0.5428 \,\text{moles}\) = \(0.092 \,\text{moles of sodium ions}\)
04

Calculate the mass of sodium chloride needed

Since sodium chloride (NaCl) dissociates into sodium ions (Na+) and chloride ions (Cl-) in a 1:1 ratio, the moles of sodium ions required will be equal to the moles of NaCl. To determine the mass of sodium chloride needed, we can use the formula: Mass of NaCl = Moles of NaCl × Molar mass of NaCl The molar mass of NaCl = 58.44 g/mol (22.99 g/mol for Na + 35.45 g/mol for Cl) Now let's calculate the mass of NaCl: Mass of NaCl = \(0.092 \,\mathrm{moles} \times 58.44 \,\frac{\mathrm{g}}{\mathrm{mol}}\) = \(5.375 \,\mathrm{g}\) So, approximately 5.375 grams of sodium chloride should be added to the blood to increase the sodium ion concentration to 0.138 M without any change in blood volume.

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Most popular questions from this chapter

A solution of \(105.0 \mathrm{~mL}\) of \(0.300 \mathrm{M} \mathrm{NaOH}\) is mixed with a solution of \(150.0 \mathrm{~mL}\) of \(0.060 \mathrm{MAlCl}_{3} .\) (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

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In each of the following pairs, indicate which has the higher concentration of \(\mathrm{Cl}^{-}\) ion: (a) \(0.10 \mathrm{MAlCl}_{3}\) solution or a $0.25 \mathrm{MLiCl}\( solution, (b) \)150 \mathrm{~mL}\(. of a \)0.05 \mathrm{MMnCl}_{3}\( solution or \)200 \mathrm{~mL}$. of \(0.10 \mathrm{M} \mathrm{KCl}\) solution, (c) a \(2.8 M H C l\) solution or a solution made by dissolving $23.5 \mathrm{~g}\( of \)\mathrm{KCl}\( in water to make \)100 \mathrm{~mL}$ of solution.

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