A 10.00mL solution of .600M NH3 is titrated with .200M HCl. Calculate the pH aft

Question

A 10.00mL solution of .600M NH3 is titrated with .200M HCl. Calculate the pH after adding 10.0mL of HCl. I used an ICE diagram to solve for equilibrium amounts and got .004mol of NH3 and .002mol of NH4+ and plugged it in to the Henderson-Hachelbach equation pH= pKa - log(base/acid) pH= 9.26-log(.004/.002)= 8.96 but this isn't right.... what am I missing?

Explanation / Answer

This is a weak base / conjugate acid problem. 10.0 mL x 0.60 M NH3 = 6 mmol NH3 Case after 10.0 mL of HCl has been added- 10.0 mL x 0.20 M = 2 mmol HCl --> 2 mmol NH4^+1 6 mmol -2 mmol -->4 mmol NH3 Total volume = 10.0 mL + 10.0 mL = 20.0 mL Kb of NH3 = 1.82x10^-5 Ka of NH4+ = 5.5x10^-10 NH3 + H2O NH4^+1 + OH^-1 Kb = [OH-][NH4+]/[NH3] [OH-] = Kb [NH3]/[NH4+] pOH = pKb - log(NH3/NH4+) pOH = 4.74 - log(4/2) = 4.74 - 0.301 = 4.439 pH = 9.561 (pH + pOH = 14.0)

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