Ex: 5 To 15 ml of 0.4mol/1 of Al_(2)(SO_(4))_(3) solution(S), 235 ml of water is added. A solution S_(1) is obtained. 1. Choose, from the following list, the appropriate material for this preparation. Available material list: * 5 mL, 10 mL and 20 mL volumetric pipets; * 5 mL, 10 mL and 20 mL graduated pipets; * 50 mL, 100 mL and 200 mL beakers; * 50 mL, 200 mL and 250 mL volumetric flasks; A pipet filler. 2. Calculate the molar concentration of the solution obtained. 3. Deduce the molar concentration of the ionic species presents in the solution S_(1) Aluminum sulfate solution

1. To prepare the solution, you would need a 20 mL volumetric flask to accurately measure and transfer the 15 mL of $0.4mol/1$ $Al_{2}(SO_{4})_{3}$ solution. Additionally, a pipet filler would be useful for measuring the exact volume of the solution.<br /><br />2. To calculate the molar concentration of the solution obtained, you can use the formula:<br /><br />\[C_1V_1 = C_2V_2\]<br /><br />where:<br />- \(C_1\) is the initial concentration of the solution (0.4 mol/L)<br />- \(V_1\) is the initial volume of the solution (15 mL)<br />- \(C_2\) is the final concentration of the solution (unknown)<br />- \(V_2\) is the final volume of the solution (15 mL + 235 mL = 250 mL)<br /><br />Rearranging the formula to solve for \(C_2\):<br /><br />\[C_2 = \frac{C_1V_1}{V_2}\]<br /><br />Substituting the given values:<br /><br />\[C_2 = \frac{(0.4 \, \text{mol/L})(15 \, \text{mL})}{250 \, \text{mL}}\]<br /><br />\[C_2 = \frac{6 \, \text{mol}}{250 \, \text{mL}}\]<br /><br />\[C_2 = 0.024 \, \text{mol/L}\]<br /><br />Therefore, the molar concentration of the solution obtained is 0.024 mol/L.<br /><br />3. To deduce the molar concentration of the ionic species present in the solution $S_{1}$, we need to consider the dissociation of $Al_{2}(SO_{4})_{3}$ in water. $Al_{2}(SO_{4})_{3}$ dissociates into 2 aluminum ions ($Al^{3+}$) and 3 sulfate ions ($SO_{4}^{2-}$) per formula unit. Therefore, the molar concentration of each ionic species will be:<br /><br />For $Al^{3+}$:<br /><br />\[C_{Al^{3+}} = 2 \times C_{Al_{2}(SO_{4})_{3}} = 2 \times 0.024 \, \text{mol/L} = 0.048 \, \text{mol/L}\]<br /><br />For $SO_{4}^{2-}$:<br /><br />\[C_{SO_{4}^{2-}} = 3 \times C_{Al_{2}(SO_{4})_{3}} = 3 \times 0.024 \, \text{mol/L} = 0.072 \, \text{mol/L}\]<br /><br />Therefore, the molar concentration of the ionic species present in the solution $S_{1}$ is 0.048 mol/L for $Al^{3+}$ and 0.072 mol/L for $SO_{4}^{2-}$.