The pH of a buffer solution that is 0.50 M in NH_3 and 0.20 M in NH_4Cl is 9.65.

Question

The pH of a buffer solution that is 0.50 M in NH_3 and 0.20 M in NH_4Cl is 9.65. If 0.0150 mole of solid NaOH is added to 500.0 mL of the buffer, then which statement best explains what happened to the pH of the system? A. NaOH reacts with NH_3 to produce more NH_4Cl. The ratio of NH_3 to NH_4Cl in the Henderson-Hasselbalch equation increases, so the pH increases slightly. B. The pH of the system increases significantly because strong base is added C. NaOH reacts with NH_4Cl_2 to produce more NH_3. The ratio of NH_3 to NH_4Cl in the Henderson-Hasselbalch equation decreases, so the pH decreases slightly D. NaOH reacts with NH_3 to produce more NH_4Cl. The ratio of NH_3 to NH_4Cl in the Henderson-Hasselbalch equation decreases, so the pH decreases slightly E. NaOH reacts with NH_4Cl to produce more NH_3. The ratio of NH_3 to NH_4Cl in the Henderson-Hasselbalch equation increases, so the pH increases slightly F. The pH of the system does not change because it is a buffer G. The pH of the system decreases significantly because strong acid is added

Explanation / Answer

correcct statement : E)

millimoles of NH3 = 0.5 x 500 = 250

millimoles of NH4Cl = 0.20 x 500 = 100

millimoles of NaOH added C = 15

pOH = pKb + log [salt - C / base + C]

        = 4.75 + log [100 - 15 / 250 + 15]

        = 4.26

pH = 9.74

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