let molarity of HCl = x
[aceticacid]=2+x
[acetate]=1-x
PHÂ Â = Pka + log[acetate]/ [aceticacid]
3.73Â Â = 4.75 + log1-x/2+x
log1-x/2+x = 3.73-4.75
log1-x/2+x  = -1.02
1-x/2+x      =
10-1.02
1-x/2+x     = 0.0955
1-x         Â
= 0.0955*(2+x)
x         =
0.74
molarity of HCl  = x = 0.74 M
no of moles of HCl  = molarity * volume In L
                              Â
= 0.74*1 = 0.74 moles
mass of
HCl          Â
= no of moles * gram molar mass
                             Â
= 0.74*36.5Â Â = 27.01g of HCl >>>>>
answer
