What is the pH of 0.256 M ammonium chloride NH4Cl?

Determine the volume (in mL) of 2.07 M potassium hydroxide (KOH) that must be added to 142 mL of 0.137 M ammon?

• Answer:

  The Henderson-Hasselbalch equation for buffers is pH = pKa + log (moles base / moles acid) In this problem, moles of base = moles of NH3 and moles of acid = moles of NH4+ 9.95 = 9.25 + log (moles NH3 / moles NH4+) 0.70 = log (moles NH3 / moles NH4+) 10^0.70 = 5.01 = moles NH3 / moles NH4+ So we need a 5:1 ratio of NH3 to NH4+. To produce NH3, we react NH4Cl with KOH: NH4+ + OH- ==> NH3 + H2O initial moles NH4+ = M NH4+ x L NH4+ = (0.137)(0.142) = 0.0195 moles NH4+ Let x = moles of NH3 needed and y = moles of NH4+ needed. Then x + y = 0.0195 since NH4+ and NH3 are in a 1:1 mole ratio in the reaction above. Also, x / y = 5.01. So x = 5.01y. Substitute that into x + y = 0.0195. x + y = 0.0195 5.01y + y = 0.0195 6.01y = 0.0195 y = 0.0195 / 6.01 = 0.00324 moles NH4+ x = 5.01y = 5.01 x 0.00324 = 0.0162 moles NH3 To produce 0.0162 moles of NH3 in the above reaction, we have to add 0.0162 moles of OH-. moles OH- = M OH- x L OH- 0.0162 = (2.07) x L OH- 0.0162 / 2.07 = L OH- = 0.00783 = 7.83 mL 2.07 M KOH Check my math!