Question

An IV solution contains 135 mEq/L Na+, 45 mEq/L K+, and 12 mEq/L Ca2+. If HCO3- is the only anion in the solution, what is its concentration in mEq/L?

Answer 1

The concentration of HCO3- in the IV solution, accounting for the 2+ charge of Ca2+, is 204 mEq/L. This is because the total positive charge, including Na+, K+, and Ca2+, adds up to 204 mEq/L, which must be balanced by an equivalent total negative charge from HCO3- for the solution to be electrically neutral.

Step-by-step explanation:

The student is asking about the concentration of HCO3- in an IV solution given the concentrations of various cations. In order for the solution to be electrically neutral, the sum of the positive charges (cations) must equal the sum of the negative charges (anions). The IV solution contains 135 mEq/L Na+, 45 mEq/L K+, and 12 mEq/L Ca2+. To find the concentration of HCO3-, we must first account for the fact that Ca2+ has a 2+ charge, contributing double the mEq for each mole compared to Na+ and K+.

Let's calculate the total positive charge:

• Na+ = 135 mEq/L
• K+ = 45 mEq/L
• Ca2+ = 12 mEq/L × 2 = 24 mEq/L

Total positive charge = 135 mEq/L + 45 mEq/L + 24 mEq/L = 204 mEq/L

The total negative charge provided by HCO3- has to be equal to the total positive charge. Hence, the concentration of HCO3- in the solution is 204 mEq/L.