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