A 100.0 mL sample of 0.18 mol L^-1 HClO₄ is titrated with 0.27 mol L^-1 NaOH. Determine the pH of the solution after the addition of 30.0 mL of NaOH. A) 0.86 B) 1.21 C) 2.00 D) 1.12 E) 2.86

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The correct answer is D.Explanation:The reaction between HClO4 and NaOH is:HClO4 + NaOH -> NaClO4 + H2OThe moles of HClO4 initially present are:0.18 mol L-1 * 0.100 L = 0.018 molThe moles of NaOH added are:0.27 mol L-1 * 0.030 L = 0.0081 molThe moles of HClO4 remaining are:0.018 mol - 0.0081 mol = 0.0099 molThe moles of NaClO4 formed are:0.0081 molThe concentration of NaClO4 is:0.0081 mol / 0.130 L = 0.0623 mol L-1The pH of the solution is:pH = -log[H+][H+] = Kw / [OH-][OH-] = 0.27 mol L-1 * 0.030 L / 0.130 L = 0.0623 mol L-1[H+] = 1.0 * 10^-14 / 0.0623 mol L-1 = 1.61 * 10^-13 mol L-1pH = -log(1.61 * 10^-13) = 12.79The pH of the solution after the addition of 30.0 mL of NaOH is 1.12.

A 1000 ML Sample of 0