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OpenStudy (anonymous):

Indicate the concentration of each ion present in the solution formed by mixing 2.38 g of NaCl in 50.0 mL of 0.400 M CaCl2 solution.

OpenStudy (anonymous):

The common ion of Cl increased [NaCl] = (2.38g / (58.44g/mol)) / (0.050L) = 0.81M NaCl ---> Na+ + Cl- [Na] = 0.81M [Cl] = 0.81M CaCl2 ----> Ca2+ + 2Cl- [Ca] = 0.4M [Cl] = 2 * 0.4 = 0.8M final [Na] = 0.81M [Ca] = 0.4 M [Cl] = 0.81+0.8 = 1.61M

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