What is the pH of the resulting solution if 40.mL of 0.432 M methylamine,CH₃NH₂,is added to 15 mL of 0.234 M HCl? Assume that the volumes of the solutions are additive.K_a = 2.70 × 10^-11for CH₃NH₃⁺. A)2.84 B)4.02 C)9.97 D)11.16

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The reaction between methylamine (a weak base) and HCl (a strong acid) will result in the formation of the methylammonium ion, CH3NH3+, and Cl-. To find the pH, we first need to determine the concentration of the methylammonium ion formed. Moles of CH3NH2 = 0.040 L * 0.432 M = 0.01728 molesMoles of HCl = 0.015 L * 0.234 M = 0.00351 molesSince HCl is limiting, it will react completely with CH3NH2, leaving some unreacted CH3NH2.Moles of CH3NH3+ formed = moles of HCl = 0.00351 molesMoles of unreacted CH3NH2 = 0.01728 - 0.00351 = 0.01377 molesTotal volume = 40 mL + 15 mL = 55 mL = 0.055 LConcentration of CH3NH3+ = 0.00351 moles / 0.055 L = 0.0638 MConcentration of CH3NH2 = 0.01377 moles / 0.055 L = 0.2504 MUsing the Henderson-Hasselbalch equation:pH = pKa + log([base]/[acid])Given pKa = -log(2.70 × 10^-11) = 10.57pH = 10.57 + log(0.2504/0.0638) = 10.57 + log(3.925) = 10.57 + 0.594 = 11.16

What Is the PH of the Resulting Solution If 40