pKa of the acid = 4.2
Amount of the base required for neutralization = 20 mL
so,
a)let the amount of dissociation be x.so,
6.3*10^-5 = x^2/(0.283-x)
or x=0.00419
so pH=2.377
b)a buffer solution will be formed.
so, ph=pKa+log(salt/acid)
=4.2 + log(10*0.283/(20*0.283-10*0.283))
=4.2
c)
a buffer solution will be formed.
so, ph=pKa+log(salt/acid)
=4.2 + log(15*0.283/(20*0.283-15*0.283))
=4.677
d)
a buffer solution will be formed.
so, ph=pKa+log(salt/acid)
=4.2 + log(19*0.283/(20*0.283-19*0.283))
=5.478
e)
a buffer solution will be formed.
so, ph=pKa+log(salt/acid)
=4.2 + log(19,95*0.283/(20*0.283-19.95*0.283))
=6.8
f)concentration of the salt formed = 0.283/2
=0.1415
so,
since this is a salt of a weak acid and a strong base,
pH=7 + 0.5pKa+0.5*log(C)
=7+0.5*4.2 + 0.5*log(0.1415)
=8.675
g) there is excess of base present.
[OH-]=0.05*0.285/(20+20.05)
=3.55*10^-4
so pH=10.55
f)
there is excess of base present.
[OH-]=5*0.285/(20+25)
=0.031667
so pH=12.5
