concentration of base=(1.921 g
/59.11g/mol)/0.050L=0.0324mol//0.050L=0.6499M
B + H2O <--> BH^+1 + OH^-1
Kb = [BH+][OH-]/[B]
[BH+] = [OH-]=x
Kb = 6.3 x 10-5 =x2/0.6499
solving for x
x=6.39x10^-3
pOH=-log(6.39x10^-3)=2.19
pH=14-2.19=11.8= pH of the original amine solution
b,no of moles of HCl=0.10molx0.1625L=0.01625mol
no of moles of base=0.0324mol
0.01625mol of HCl reacts with 0.01625mol of base gives0.01625mol
of BH+
remaining base=0.0324mol-0.01625mol=0.01615mol
new concentration of BH+ =0.01625mol /0.2125L=0.076M
concentration of base in new
volume=0.01615mol//0.2125L=0.076M
pOH=pKb+log(0.076M/0.076M)
pOH=4.20+0
pH=14-4.20=9.79
c,no of moles of HCl=0.10molx0.325L=0.0325mol
no of moles of base=0.03249mol
all the base is reacts with the acid
so BH+=0.0325mol/0.375L=0.086M
acid dissociation constant,ka=kw/kb=1x10^-14/ 6.3 x
10-5 =1.58x10^-10
ka=x2/0.086M-x=1.58x10^-10
solving for x
x=0.00000367=[BH+]
pH=5.43
