A solution is prepared by mixing 200 ml of 0.1M NH3 and 300 mL of 0.05 NH4Cl. Water is added to bring total solution volume to 750 mL. Kb for nH3 is 1.8x10^-5
please solve, the answer is 9.4 Please do it without the use of Handerson's equation and apply more of an ice table approach. Thank you
Moles of NH3 = molarity * volume(in litre)
= 0.200*0.1 = 0.02
Moles of NH4Cl = 0.300*0.05 = 0.015
pOh = pKb + log(NH4Cl/NH3)
pKb = -log (kB)
= -log(1.8*10^-5) = 4.74
pOH = 4.74 + log ( 0.015 / 0.02)
= 4.615
pH = 14-poh
= 14-4.615 = 9.385
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