In: Chemistry
What is the pH of an aqueous solution made by combining 44.32 mL of a 0.3616 M ammonium chloride with 36.75 mL of a 0.3087 M solution of ammonia to which 4.002 mL of a 0.0516 M solution of NaOH was added?
no of moles of NH4Cl = molarity * volume in L
= 0.3616*0.04432 = 0.016 moles
no of moles of NH3 = molarity * volume in L
= 0.3087*0.03675 = 0.0113 moles
PKb = 4.75
no of moles of NaOH = molarity * volume in L
= 0.0516*0.004002 = 0.000207 moles
no of moles of NH4l after addition of NaOH = 0.016-0.000207 = 0.0158 moles
no of moles of NH3 after addition of NaOH = 0.0113+0.000207 = 0.011507 moles
POH = PKb + log[NH4Cl]/[NH3]
= 4.75 + log0.0158/0.011507
= 4.75+0.137
= 4.887
PH = 14-POH
= 14-4.887 = 9.113 >>>>answer