Calculate the pH at the equivalence point for the titration of 50.0 mL of 0.20 M NH₃ with 0.20 M HCl. [Kb(NH₃)= 1.8 × 10^-5]

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Quizplus · Accepted Answer

At the equivalence point, moles of HCl = moles of NH₃ n(HCl) = 0.20 M x 50.0 mL = 0.010 mol n(NH₃) = 0.010 mol Therefore, the final volume is 100.0 mL and the final concentration of NH₃ is 0.10 M. NH₃ is a weak base, so at the equivalence point, half of it will have reacted with HCl to form its conjugate acid (NH2S1U1BH) and the other half will remain as the base. The reaction is: NH₃ + HCl -> NH2S1U1BH + Cl- The moles of base remaining will be equal to the moles of acid formed, which is 0.005 mol. The initial moles of base was 0.010 mol, so half of it has reacted, leaving 0.005 mol of NH₃. The final concentration of NH₃ is 0.005 mol/0.100 L = 0.050 M. Now we can set up the Kb expression for NH₃: Kb = [NH2S1U1BH][OH-]/[NH₃] 1.8 × 10-5 = x^2/(0.050 - x) Since x is much less than 0.050, we can simplify: 1.8 × 10-5 = x^2/0.050 x = 1.34 × 10-3 M Now we can find the pH: pH = 14 - pOH = 14 - (-log[OH-]) pH = 14 - (-log[1.34 × 10-3]) pH = 5.13 Therefore, the best choice is C.

Calculate the PH at the Equivalence Point for the Titration