Assuming all the nh3 dissolves and that the volume of the solution remains at 0.300 l , calculate the ph of the resulting solution.

Assume p=735 Torr V= 7.6L R=62.4 T= 295 PV-nRT (735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K) 0.30346 moles of NH3 Find moles 0.300L solution of 0.300 M HCL = 0.120 moles of HCL 0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind Find molarity 0.120 moles of NH4+/0.300L = 0.400 M NH4+ 0.18346 moles of NH3/0.300L = 0.6115 M NH3 NH4OH --> NH4 & OH- Kb = [NH4+][OH]/[NH4OH] 1.8 e-5=[0.300][OH-]/[0.6115] [OH-]=1.6e-5 pOH= 4.79 PH=9.21 .

Woah that’s a lot to read xd i’ll try though. i’ll get back t you if. i know the andwer

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