Calculate the pH for 0.001 M HCl.

You're on a roll — let's evaluate each statement to determine which are correct. Statement I: The pH of 0.001 M H₂SO₄(aq) and 0.001 M HCl(aq) are the same. Step 1: Calculate the pH for 0.001 M HCl. HCl is a strong monoprotic acid, so it dissociates completely. [H^+] = 0.001 M = 10^-3 M pH = -[H^+] = -(10^-3) = 3 Step 2: Calculate the pH for 0.001 M H₂SO₄. H₂SO₄ is a strong diprotic acid. For dilute solutions, both protons are considered to dissociate completely. H_2SO_4 2H^+ + SO_4^2- So, for 0.001 M H₂SO₄, the concentration of H^+ is approximately 2 × 0.001 M = 0.002 M = 2 × 10^-3 M. pH = -(2 × 10^-3) = 3 - (2) ≈ 3 - 0.301 = 2.699 Since 3 ≠ 2.699, the pH values are not the same. Therefore, Statement I is incorrect. Statement II: The pOH of 0.001 M NaOH(aq) is 3. Step 1: Determine the concentration of OH^-. NaOH is a strong base, so it dissociates completely. NaOH Na^+ + OH^- For 0.001 M NaOH, [OH^-] = 0.001 M = 10^-3 M. Step 2: Calculate the pOH. pOH = -[OH^-] = -(10^-3) = 3 Therefore, Statement II is correct. Statement III: The ionic product of water, Kw, is temperature dependent. The autoionization of water is an endothermic process: H_2O(l) H^+(aq) + OH^-(aq) According to Le Chatelier's principle, an increase in temperature shifts the equilibrium to the right, increasing the concentrations of both H^+ and OH^-. Since K_w = [H^+][OH^-], K_w increases with temperature. Therefore, Statement III is correct. Statement IV: The pKb of 0.010 M NH₃ solution is greater than 7. (kb of NH₃ = 1.8 x 10⁻⁵) Step 1: Calculate the pKb from the given K_b. The pKb is a constant for a given base and does not depend on its concentration. pKb = -(K_b) pKb = -(1.8 × 10^-5) pKb = 5 - (1.8) pKb ≈ 5 - 0.255 = 4.745 Since 4.745 is not greater than 7, the statement is incorrect. Therefore, Statement IV is incorrect. Conclusion: Statements II and III are correct. The final answer is II and III. Send me the next one 📸