weak base vs strong acid(S.A)
no of mol of NH3 = 40*0.3 = 12 mmol
no of mol of HCl required to reach equivalence point = 12 mmol
a) before addition of S.A
pkb of NH3 = -logkb = -log(1.8*10^-5) = 4.74
C = concentration of NH3 = 0.1 M
pOH = 1/2(pkb-logC)
= 1/2(4.74-log0.3)
= 2.63
pH = 14-2.63 = 11.37
b) afetr 20% of the total volume of HCl
no of mol of HCl added = 12*20/100 = 2.4 mmol = No of mol of NH4+
no of mol of NH3 = 40*0.3 = 12 mmol
pOH = pkb+log(NH4+/NH3)
= 4.74 + log(2.4/(12-2.4))
= 4.14
pH = 14-4.14 = 9.86
c) afetr 50% of the total volume of HCl
no of mol of HCl added = 12*50/100 = 6 mmol = No of mol of NH4+
no of mol of NH3 = 40*0.3 = 12 mmol
pOH = pkb+log(NH4+/NH3)
= 4.74 + log(6/(12-6))
= 4.74
pH = 14-4.74 = 9.26
d) afetr 95% of the total volume of HCl
no of mol of HCl added = 12*95/100 = 11.4 mmol = No of mol of NH4+
no of mol of NH3 = 40*0.3 = 12 mmol
pOH = pkb+log(NH4+/NH3)
= 4.74 + log(11.4/(12-11.4))
= 6.02
pH = 14-6.02 = 7.98
c) afetr 100% of the total volume of HCl
no of mol of HCl added = 12 mmol = No of mol of NH4cl
volume of HCl added = 12/0.5 = 24 ml
no of mol of NH3 = 40*0.3 = 12 mmol
concentration of salt(NH4Cl) = n/V
= 12/(40+24)
= 0.1875 M
pH = 7-1/2(pkb+logC)
= 7-1/2(4.74+log0.1875)
= 5
d) afetr 100% of the total volume of HCl
no of mol of HCl added = 12*105/100 = 12.6 mmol
volume of HCl added = 12.6/0.5 = 25.2 ml
no of mol of NH3 = 40*0.3 = 12 mmol
concentration of excess HCl = (12.6-12)/(25.2+40)
= 0.0092 M
pH = -log(H3O+)
= -log0.0092
= 2.04
I need help to answer the bold questions please. More so help on 100% of HCl...
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