Calculate the normality of each of the following solutions. a. 0.250\(M \mathrm{HCl}\) b. 0.105\(M \mathrm{H}_{2} \mathrm{SO}_{4}\) c. \(5.3 \times 10^{-2} M \mathrm{H}_{3} \mathrm{PO}_{4}\) d. 0.134 \(\mathrm{M} \mathrm{NaOH}\) e. 0.00521 \(\mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) What is the equivalent mass for each of the acids or bases listed above?

For each compound, count the moles of ions that can participate in the acid-base reaction, i.e., the number of H+ ions from acids and OH- ions from bases. a. \(HCl\) ⟶ 1 H+ ion; n = 1 b. \(H_2SO_4\) ⟶ 2 H+ ions; n = 2 c. \(H_3PO_4\) ⟶ 3 H+ ions; n = 3 d. \(NaOH\) ⟶ 1 OH- ion; n = 1 e. \(Ca(OH)_2\) ⟶ 2 OH- ions; n = 2

Now, to find the normality of each solution, multiply the molarity by the number of equivalents (n): a. 0.250 M \(HCl\): Normality = 0.250 × 1 = 0.250 N b. 0.105 M \(H_2SO_4\): Normality = 0.105 × 2 = 0.210 N c. \(5.3 \times 10^{-2} M H_3PO_4\): Normality = \(5.3 \times 10^{-2} × 3 = 15.9 \times 10^{-2}\) N d. 0.134 M \(NaOH\): Normality = 0.134 × 1 = 0.134 N e. 0.00521 M \(Ca(OH)_2\): Normality = 0.00521 × 2 = 0.01042 N

To find the equivalent mass, divide the molar mass of each compound by the number of equivalents (n): a. \(HCl\): Molar mass = 1 + 35.5 = 36.5 g/mol; Equivalent mass = 36.5/1 = 36.5 g/equiv b. \(H_2SO_4\): Molar mass = 2 × 1 + 32 + 4 × 16 = 98 g/mol; Equivalent mass = 98/2 = 49 g/equiv c. \(H_3PO_4\): Molar mass = 3 × 1 + 31 + 4 × 16 = 98 g/mol; Equivalent mass = 98/3 ≈ 32.67 g/equiv d. \(NaOH\): Molar mass = 23 + 16 + 1 = 40 g/mol; Equivalent mass = 40/1 = 40 g/equiv e. \(Ca(OH)_2\): Molar mass = 40 + 2 × (16 + 1) = 74 g/mol; Equivalent mass = 74/2 = 37 g/equiv So, the equivalent masses of \(HCl\), \(H_2SO_4\), \(H_3PO_4\), \(NaOH\), and \(Ca(OH)_2\) are 36.5 g/equiv, 49 g/equiv, 32.67 g/equiv, 40 g/equiv, and 37 g/equiv, respectively.