ANS:D - < 7

To find the pH of a solution of sulfuric acid (H2SO4H2​SO4​) with a 5% concentration, we first need to determine the molarity of the solution. Since the concentration is given as a percentage, we assume it refers to mass/volume percent. For a 5% solution, this means 5 grams of H2SO4H2​SO4​ per 100 milliliters of solution (assuming the density of the solution is approximately 1 g/mL). The molar mass of H2SO4H2​SO4​ is approximately 98 g/mol. So, in 100 mL of solution, there are 5 g98 g/mol×1 mol0.1 L≈0.051 mol/L98 g/mol5 g​×0.1 L1 mol​≈0.051 mol/L of H2SO4H2​SO4​. Given that H2SO4H2​SO4​ is a strong acid, it dissociates completely in solution. Since it is a diprotic acid, it can donate two protons (H+H+) per molecule. So, the concentration of H+H+ ions in this 5% solution of H2SO4H2​SO4​ is 2×0.051 mol/L=0.102 mol/L2×0.051 mol/L=0.102 mol/L. Now, let's calculate the pH of the solution: pH=−log⁡[H+]=−log⁡(0.102)pH=−log[H+]=−log(0.102) pH≈−log⁡(1×10−1)pH≈−log(1×10−1) pH≈−(−1)pH≈−(−1) pH≈1pH≈1 So, the pH of the 5% solution of H2SO4H2​SO4​ is less than 7. Therefore, the correct answer is "< 7".