A patient is prescribed lithium carbonate (Li₂CO₃) 150 mg capsules PO BID. He cannot swallow capsules and prefers a liquid. The only lithium solution available is lithium citrate (Li₃C₆H₅O₇), which contains 8 mEq Li⁺ per teaspoon.

20. How many mEq of Li⁺ is found in the lithium carbonate 150 mg capsule? (MW of Li₂CO₃ = 73.89 g/mol)

21. How many milliliters of the lithium citrate solution would deliver an equivalent dose of lithium ion? Round the volume to the nearest tenth.

The lithium carbonate 150 mg capsule contains approximately 0.00406 mEq of Li'. To deliver an equivalent dose of lithium ion, approximately 0.00251 milliliters of the lithium citrate solution would be needed.

Explanation:

To calculate the mEq of Li' in the lithium carbonate 150 mg capsule, we need to convert the mass of lithium carbonate to moles and then use the molecular weight to calculate the mEq. Here's how:

Convert the mass of lithium carbonate to moles:
150 mg * (1 g / 1000 mg) * (1 mol / 73.89 g) = 0.00203 mol
Calculate the mEq of Li':
0.00203 mol * (2 mEq / 1 mol) = 0.00406 mEq

Therefore, the lithium carbonate 150 mg capsule contains approximately 0.00406 mEq of Li'.

To calculate the volume of the lithium citrate solution that would deliver an equivalent dose of lithium ion, we need to use the conversion factor between lithium citrate and lithium ion. Here's how:

Calculate the volume of lithium citrate solution:
0.00406 mEq * (1 teaspoon / 8 mEq) = 0.00050825 teaspoons
Convert teaspoons to milliliters:
0.00050825 teaspoons * (4.92892 mL / 1 teaspoon) = 0.00251 mL

Therefore, approximately 0.00251 milliliters of the lithium citrate solution would deliver an equivalent dose of lithium ion.

Learn more about calculating meq of lithium in different forms here:

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