Question

A solution contains the following ions: K⁺(13.2 mEq/L​), Mg²⁺(15.8 mEq/L​), and Sr²⁺(3.5 mEq/L​), if HCO3₃ is the only anion present in the solution, what is its concentration in LmEq​ ? Round your answer to 3 significant figures.

Answer 1

The concentration of the anion HCO3⁻, needed to balance the charges from the cations K⁺, Mg⁻², and Sr²⁻ in the solution, is 51.8 mEq/L.

Step-by-step explanation:

To calculate the concentration of HCO3⁻ as the only anion present in the solution, we need to ensure that the charges from the cations balance out the charges from the anions. We'll add up the positive charges from K+, Mg2+, and Sr2+, and then determine the concentration of HCO3⁻ needed to balance the solution electrically.

The sum of cation charges is:

• K+: 13.2 mEq/L×1 = 13.2 mEq/L
• Mg2+: 15.8 mEq/L×2 = 31.6 mEq/L
• Sr2+: 3.5 mEq/L×2 = 7.0 mEq/L

Total positive charge = 13.2 mEq/L + 31.6 mEq/L + 7.0 mEq/L = 51.8 mEq/L

Since K+ and HCO3⁻ are monovalent, and Mg2+ and Sr2+ are divalent, one mole of HCO3⁻ will balance one equivalent of a monovalent ion, and 0.5 moles will balance one equivalent of a divalent ion. To maintain electrical neutrality, the total concentration of the anion HCO3⁻ (as mEq/L) will be equal to the total positive charge from the cations, which is 51.8 mEq/L.

So, the concentration of HCO3⁻ in mEq/L is 51.8 mEq/L, rounded to three significant figures.