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What is the total anion concentration (in meq/l) of a solution that contains 6.0 meq/lna+ , 11.0 meq/lca2+ , and 3.0 meq/lli+ ?

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Answer: The total anion concentration = 6.0 + 2x12.0 + 3.0 = 33.0mEq/L

We know that each equivalent of Na+ will be combined with one equivalent of anion concentration

Each equivalent of Ca+2 will combine with one equivalent of anion

each equivalent of Li+ will combine with one equivalent of anion

As we are given with 6.0 meq/lna+ , 11.0 meq/lca2+ , and 3.0 meq/lli+ ?

So for each  6.0 meq/l  Na+, the anion concentration will be 6.0 meq/l

So for each  11 meq/l Ca+2 , the anion concentration will be 11meq/l

So for each  3.0 meq/l Li+ , the anion concentration will be 3 meq/l

Thus total anion concentration = 6 + 11 + 3 = 20 meq/l