Question

1.921 g of the weak base trimethylamine, (CH₃)₃N (MW = 59.11g/mol) (K_b = 6.3 x 10^-5), is diluted in water to give 50.0 mL of solution. This solution is titrated with 0.10 M HCl. (assume all solutions at 25⁰C)

(a) What is the pH of the original amine solution?

(b) What is the pH of the solution after 162.5 ml of HCl is titrated?

(c) What is the pH after 325ml of HCl is titrated?

Answer 1

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
K_b = 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