What is the total anion concentration (in mEq/L ) of a solution that contains 4.0mEq/LNa + , 13.0mEq/LCa 2+ , and 3.0mEq/LLi + ?
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To calculate the total anion concentration (in mEq/L) of a solution, you need to consider the charges of the cations and balance them with the corresponding anions. In this case, we have Na+, Ca2+, and Li+ as cations.
First, let's find the anions associated with each cation:
1. Sodium (Na+) has a charge of +1. The corresponding anion is chloride (Cl-), which also carries a charge of -1. Therefore, for every Na+ cation present, there is one chloride anion (Cl-) to balance the charge.
2. Calcium (Ca2+) has a charge of +2. The corresponding anion is hydroxide (OH-), which carries a charge of -1. Since the charge of the calcium ion is +2, two OH- ions are required to balance its charge.
3. Lithium (Li+) has a charge of +1. The corresponding anion is also chloride (Cl-), which has a charge of -1. Thus, for every Li+ cation, one Cl- ion is needed to balance the charge.
Now, let's calculate the total anion concentration:
- Na+ has a concentration of 4.0 mEq/L. Since Na+ requires one Cl- anion to balance its charge, the anion concentration is also 4.0 mEq/L.
- Ca2+ has a concentration of 13.0 mEq/L. Since two OH- anions are needed for each Ca2+ ion, we multiply the Ca2+ concentration by 2 to get the anion concentration: 13.0 mEq/L * 2 = 26.0 mEq/L.
- Li+ has a concentration of 3.0 mEq/L. As Li+ requires one Cl- anion for balance, the anion concentration is also 3.0 mEq/L.
Finally, we add up the anion concentrations:
4.0 mEq/L (Cl-) + 26.0 mEq/L (OH-) + 3.0 mEq/L (Cl-) = 33.0 mEq/L.
Therefore, the total anion concentration of the solution is 33.0 mEq/L.