Question

# Consider the titration of 30.0 mL of 0.050 MNH3 with 0.025 MHCl. Calculate the pH after...

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

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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