Indicate the concentration of each ion present in the solution formed by mixing (a) 42.0 \(\mathrm{mL}\) of 0.170 \(\mathrm{M} \mathrm{NaOH}\) with 37.6 \(\mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{NaOH},(\mathbf{b}) 44.0 \mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) with 25.0 \(\mathrm{mL}\) of \(0.150 \mathrm{M} \mathrm{KCl},\) (c) 3.60 \(\mathrm{g} \mathrm{KCl}\) in 75.0 \(\mathrm{mL}\) of 0.250\(M \mathrm{CaCl}_{2}\) solution. Assume that the volumes are additive.

Here, we need to apply formula of molarity, which is Moles= Molarity x Volume(L). Applying the formula: Moles of NaOH = \(0.170 \, \mathrm{M} \times 0.042 \, \mathrm{L} = 0.00714 \, \mathrm{moles}\)

From the second solution, moles of NaOH = \(0.400 \, \mathrm{M} \times 0.0376 \, \mathrm{L} = 0.01504 \, \mathrm{moles}\)

Total moles of NaOH = \(0.00714 + 0.01504 = 0.02218 \, \mathrm{moles}\) **Step 2: Adding Up the Volumes**

Total volume= \(.042 L + .0376 L = .0796 L\) **Step 3: Calculating the Concentration**

The concentration of NaOH is obtained by dividing the total number of moles of NaOH by the total volume of the solution. So, the concentration of NaOH = \( \frac{0.02218}{.0796} \, \mathrm{M} = 0.279 \, \mathrm{M}\). ##Part (b)## Given solutions are \(44.0 \, \mathrm{mL}\) of \(0.100 \, \mathrm{M} \, \mathrm{Na}_{2} \, \mathrm{SO}_{4}\) and \(25.0 \, \mathrm{mL}\) of \(0.150 \, \mathrm{M} \, \mathrm{KCl}\) Repeat steps 1 to 3 for this part, remembering that for every mole of \(Na_{2}SO_{4}\) you get 2 moles of \(Na^{+}\) ions and for every mole of \(KCL\) you get 1 mole of \(K^{+}\) ions. ##Part (c)## Given solutions are \(3.60 \, \mathrm{g}\) \(KCl\) in \(75.0 \, \mathrm{mL}\) of \(0.250 \, M \, CaCl_{2}\) solution. Repeat steps 1 to 3 for this part, remembering that for every gram of \(KCL\) you get \(\frac{3.60 \, g}{74.55 \, g/mol}=0.0483 \, moles\) of \(KCl\), which corresponds to 0.0483 moles of \(K^{+}\) ions. For every mole of \(CaCl_{2}\) you get 2 moles of \(Cl^{-}\) ions.