Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.100 mole of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(100.0 \mathrm{mL}\) of solution b. 2.5 moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in 1.25 L of solution c. \(5.00 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(500.0 \mathrm{mL}\) of solution d. \(1.00 \mathrm{g} \mathrm{K}_{3} \mathrm{PO}_{4}\) in \(250.0 \mathrm{mL}\) of solution

First, we need to find the molarity of the solution for each substance: a. Molarity \(a= \frac{0.100 \, \text{mol of} \, \mathrm{Ca(NO}_{3})_{2}}{100.0 \, \text{mL}} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} =0.00100\, \mathrm{M}\) b. Molarity \(b= \frac{2.5 \, \text{mol of} \, \mathrm{Na}_{2}\mathrm{SO}_{4}}{1.25 \, \text{L}} =2.0 \, \mathrm{M}\) c. First, we need to find the moles of \(\mathrm{NH}_{4}\mathrm{Cl}\): Moles \(c = \frac{5.00\, \mathrm{g}}{53.49 \, \mathrm{g/mol}} \approx 0.0935\, \mathrm{mol}\) Molarity \(c= \frac{0.0935 \, \text{mol of} \, \mathrm{NH}_{4}\mathrm{Cl}}{0.500 \, \text{L}} =0.187 \, \mathrm{M}\) d. First, we need to find the moles of \(\mathrm{K}_{3}\mathrm{PO}_{4}\): Moles \(d = \frac{1.00\, \mathrm{g}}{212.28 \, \mathrm{g/mol}} \approx 0.00471\, \mathrm{mol}\) Molarity \(d= \frac{0.00471 \, \text{mol of} \, \mathrm{K}_{3}\mathrm{PO}_{4}}{0.25 \, \text{L}} =0.0188 \, \mathrm{M}\)

Now, we will calculate the molarity of the resulting ions in each solution: a. \(\mathrm{Ca(NO}_{3})_{2}\) dissociates into \(\mathrm{Ca^{2+}}\) and \(\mathrm{NO_3^-}\), with a stoichiometry of 1:2. \(\mathrm{M(Ca^{2+})} = 0.00100\, \mathrm{M}\) \(\mathrm{M(NO_3^{-})} = 2 \times 0.00100\, \mathrm{M} = 0.00200\, \mathrm{M}\) b. \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) dissociates into \(\mathrm{Na^{+}}\) and \(\mathrm{SO_4^{2-}}\), with a stoichiometry of 2:1. \(\mathrm{M(Na^{+})} = 2 \times 2.0\, \mathrm{M} = 4.0\, \mathrm{M}\) \(\mathrm{M(SO_4^{2-})} = 2.0\, \mathrm{M}\) c. \(\mathrm{NH}_{4}\mathrm{Cl}\) dissociates into \(\mathrm{NH_4^{+}}\) and \(\mathrm{Cl^-}\), with a stoichiometry of 1:1. \(\mathrm{M(NH_4^{+})} = 0.187\, \mathrm{M}\) \(\mathrm{M(Cl^{-})} = 0.187\, \mathrm{M}\) d. \(\mathrm{K}_{3}\mathrm{PO}_{4}\) dissociates into \(\mathrm{K^{+}}\) and \(\mathrm{PO_4^{3-}}\), with a stoichiometry of 3:1. \(\mathrm{M(K^{+})} = 3\times 0.0188\, \mathrm{M} = 0.0564\, \mathrm{M}\) \(\mathrm{M(PO_4^{3-})} = 0.0188\, \mathrm{M}\) Final concentrations of ions: a. \(\mathrm{Ca^{2+}}: 0.00100\, \mathrm{M}\), \(\mathrm{NO_3^{-}}: 0.00200\, \mathrm{M}\) b. \(\mathrm{Na^{+}}: 4.0\, \mathrm{M}\), \(\mathrm{SO_4^{2-}}: 2.0\, \mathrm{M}\) c. \(\mathrm{NH_4^{+}}: 0.187\, \mathrm{M}\), \(\mathrm{Cl^{-}}: 0.187\, \mathrm{M}\) d. \(\mathrm{K^{+}}: 0.0564\, \mathrm{M}\), \(\mathrm{PO_4^{3-}}: 0.0188\, \mathrm{M}\)