Question

**Consider the titration of 30.0 mL of 0.050 MNH3
with 0.025 MHCl. Calculate the pH after the following
volumes of titrant have been added.**

**A.** 0 mL

**B.** 20.0 mL

**C**. 59.2 mL

**D**. 60.0 mL

**E.** 72.5 mL

**F.** 73.1

**Express your answer using two decimal
places.**

Answer #1

Kb=4.76*10^-5

A) when no HCl is added,

let the dissociation be x.so,

4.76*10^-5 = x^2/(0.05-x)

or x=0.0015 M

so pOH=-log(0.0015)

=2.82

so pH=14-2.82

=11.18

B) amount of HCl needed for neutralization be x.so,

x*0.025 = 30*0.05

=60 mL

b) when 20 mL is added,

pOH=pKb+log(salt//acid)

= 4.76 + log(20*0.025/(30*0.05 - 0.025*20))

=4.46

so pH=14-4.46

=9.54

C) pOH=pKb+log(salt/acid)

= 4.76 + log(59.2*0.025/(30*0.05 - 0.025*59.2))

=4.46

so pH=14-4.46

=6.63

D) now there is no moles NH3 and HCl

the concentration of NH4+ ion = (30*0.05)/(60+30)

=0.01667 M

so pH = 7 - 0.5pKb -0.5*log(C)

=7 - 0.5*4.76 + 0.5*0.01667

=4.63

E) now excess of the HCl is added,

[H+] = 72.5*0.025/(72.5+30)

=0.0177

pH=-log(0.01768)

=1.7525

F)[H+] = 73.1*0.025/(73.1+30)

= -0.01772

so pH=-log(0.01772)

= 1.75

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