What is the pH of the solution containing 0.20 M NH3 and 0.15M NH4Cl?
Kb (NH3)= 1.8*10-5 and pKb = 4.74
pOH = pKb - log ([NH4Cl] / [NH3] )
pOH = 4.74 - log (0.15/0.2)
pOH = 4.74 - log 0.75
pOH = 4.74 + 0.12
pOH = 4.86
pH = 14 - pOH
pH = 14.00 - 4.86
pH = 9.14
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